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Commita86931f

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author
luzhipeng
committed

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25 files changed

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25 files changed

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‎README.md

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@@ -82,14 +82,17 @@ leetcode 题解，记录自己的 leetcode 解题之路。
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####简单难度
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85-
-[0020.Valid Parentheses](./problems/20.validParentheses.md) 🖊
85+
-[0020.Valid Parentheses](./problems/20.validParentheses.md)
8686
-[0026.remove-duplicates-from-sorted-array](./problems/26.remove-duplicates-from-sorted-array.md)
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-[0088.merge-sorted-array](./problems/88.merge-sorted-array.md)
88+
-[0121.best-time-to-buy-and-sell-stock](./problems/121.best-time-to-buy-and-sell-stock.md) 🆕
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-[0122.best-time-to-buy-and-sell-stock-ii](./problems/122.best-time-to-buy-and-sell-stock-ii.md) 🆕
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-[0136.single-number](./problems/136.single-number.md)
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-[0167.two-sum-ii-input-array-is-sorted](./problems/167.two-sum-ii-input-array-is-sorted.md)
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-[0169.majority-element](./problems/169.majority-element.md)
9193
-[0190.reverse-bits](./problems/190.reverse-bits.md)
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-[0191.number-of-1-bits](./problems/191.number-of-1-bits.md)
95+
-[0198.house-robber.md](./problems/198.house-robber.md) 🆕
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-[0203.remove-linked-list-elements](./problems/203.remove-linked-list-elements.md)
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-[0206.reverse-linked-list](./problems/206.reverse-linked-list.md)
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-[0219.contains-duplicate-ii](./problems/219.contains-duplicate-ii.md)
@@ -115,7 +118,8 @@ leetcode 题解，记录自己的 leetcode 解题之路。
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-[0075.sort-colors](./problems/75.sort-colors.md)
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-[0078.subsets](./problems/78.subsets.md) 🆕
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-[0086.partition-list](./problems/86.partition-list.md)
118-
-[0090.subsets-ii](./problems/90.subsets-ii.md) 🆕
121+
-[0090.subsets-ii](./problems/90.subsets-ii.md)
122+
-[0091.decode-ways](./problems/91.decode-ways.md) 🆕
119123
-[0092.reverse-linked-list-ii](./problems/92.reverse-linked-list-ii.md)
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-[0094.binary-tree-inorder-traversal](./problems/94.binary-tree-inorder-traversal.md)
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-[0102.binary-tree-level-order-traversal](./problems/102.binary-tree-level-order-traversal.md)
@@ -131,13 +135,15 @@ leetcode 题解，记录自己的 leetcode 解题之路。
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-[0236.lowest-common-ancestor-of-a-binary-tree](./problems/236.lowest-common-ancestor-of-a-binary-tree.md)🆕
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-[0238.product-of-array-except-self](./problems/238.product-of-array-except-self.md) 🆕
133137
-[0240.search-a-2-d-matrix-ii](./problems/240.search-a-2-d-matrix-ii.md)
134-
-[0279.perfect-squares](./problems/279.perfect-squares.md)🖊
138+
-[0279.perfect-squares](./problems/279.perfect-squares.md)
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-[0309.best-time-to-buy-and-sell-stock-with-cooldown](./problems/309.best-time-to-buy-and-sell-stock-with-cooldown.md) 🆕
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-[0322.coin-change](./problems/322.coin-change.md)
136141
-[0334.increasing-triplet-subsequence](./problems/334.increasing-triplet-subsequence.md)
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-[0328.odd-even-linked-list](./problems/328.odd-even-linked-list.md)
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-[0416.partition-equal-subset-sum](./problems/416.partition-equal-subset-sum.md)
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-[0445.add-two-numbers-ii](./problems/445.add-two-numbers-ii.md)
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-[0454.4-sum-ii](./problems/454.4-sum-ii.md) 🆕
146+
-[0494.target-sum](./problems/494.target-sum.md) 🆕
141147
-[0518.coin-change-2](./problems/518.coin-change-2.md)
142148
-[0875.koko-eating-bananas](./problems/875.koko-eating-bananas.md)
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-[0877.stone-game](./problems/877.stone-game.md)
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<mxfile modified="2019-05-04T06:00:45.801Z" host="www.draw.io" agent="Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/73.0.3683.86 Safari/537.36" etag="8lZpRP_onmLUlqddJt9q" version="10.6.6" type="device"><diagram id="YQNoRghu4fKXr0qcQk76" name="第 1 页">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</diagram></mxfile>
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‎assets/drawio/198.house-robber.drawio

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‎assets/drawio/494.target-sum.drawio

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/*
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*@lc app=leetcode id=108 lang=javascript
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*
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* [108] Convert Sorted Array to Binary Search Tree
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*
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* https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/description/
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*
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* algorithms
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* Easy (49.37%)
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* Total Accepted: 255.2K
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* Total Submissions: 507.2K
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* Testcase Example: '[-10,-3,0,5,9]'
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*
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* Given an array where elements are sorted in ascending order, convert it to a
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* height balanced BST.
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*
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* For this problem, a height-balanced binary tree is defined as a binary tree
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* in which the depth of the two subtrees of every node never differ by more
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* than 1.
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*
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* Example:
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*
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*
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* Given the sorted array: [-10,-3,0,5,9],
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*
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* One possible answer is: [0,-3,9,-10,null,5], which represents the following
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* height balanced BST:
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*
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* ⁠ 0
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* ⁠ / \
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* ⁠ -3 9
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* ⁠ / /
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* ⁠-10 5
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*
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*
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*/
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/**
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* Definition for a binary tree node.
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* function TreeNode(val) {
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* this.val = val;
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* this.left = this.right = null;
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* }
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*/
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/**
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*@param {number[]} nums
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*@return {TreeNode}
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*/
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varsortedArrayToBST=function(nums){
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// 由于数组是排序好的，因此一个思路就是将数组分成两半，一半是左子树，另一半是右子树
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// 然后运用“树的递归性质”递归完成操作即可。
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if(nums.length===0)returnnull;
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constmid=nums.length>>1;
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constroot=newTreeNode(nums[mid]);
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root.left=sortedArrayToBST(nums.slice(0,mid));
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root.right=sortedArrayToBST(nums.slice(mid+1))
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returnroot;
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// 扩展： 这道题启示我们如果是一个非排序的数组，我们可以先进行排序然后再按上述思路进行。
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};
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‎backlog/202.happy-number.js

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/*
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*@lc app=leetcode id=202 lang=javascript
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*
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* [202] Happy Number
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*
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* https://leetcode.com/problems/happy-number/description/
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*
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* algorithms
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* Easy (44.36%)
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* Total Accepted: 227.2K
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* Total Submissions: 505.7K
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* Testcase Example: '19'
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*
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* Write an algorithm to determine if a number is "happy".
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*
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* A happy number is a number defined by the following process: Starting with
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* any positive integer, replace the number by the sum of the squares of its
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* digits, and repeat the process until the number equals 1 (where it will
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* stay), or it loops endlessly in a cycle which does not include 1. Those
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* numbers for which this process ends in 1 are happy numbers.
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*
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* Example: 
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*
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*
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* Input: 19
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* Output: true
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* Explanation:
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* 1^2 + 9^2 = 82
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* 8^2 + 2^2 = 68
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* 6^2 + 8^2 = 100
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* 1^2 + 0^2 + 0^2 = 1
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*
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*/
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functionsquareSum(n){
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letsum=0,tmp;
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while(n){
37+
tmp=n%10;
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sum+=tmp*tmp;
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n=Math.floor(n/10);
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}
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returnsum;
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}
43+
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functionisHappyWithMapper(n,visited){
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if(n===1)returntrue;
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if(visited[n])returnfalse;
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visited[n]=true;
48+
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returnisHappyWithMapper(squareSum(n),visited);
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}
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/**
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*@param {number} n
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*@return {boolean}
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*/
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varisHappy=function(n){
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constvisited={};
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returnisHappyWithMapper(n,visited);
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};
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‎backlog/204.count-primes.js

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/*
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*@lc app=leetcode id=204 lang=javascript
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*
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* [204] Count Primes
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*
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* https://leetcode.com/problems/count-primes/description/
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*
8+
* algorithms
9+
* Easy (28.33%)
10+
* Total Accepted: 229.8K
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* Total Submissions: 798.7K
12+
* Testcase Example: '10'
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*
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* Count the number of prime numbers less than a non-negative number, n.
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*
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* Example:
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*
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*
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* Input: 10
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* Output: 4
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* Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
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*
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*
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*/
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/**
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*@param {number} n
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*@return {number}
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*/
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varcountPrimes=function(n){
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// tag: 数论
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// if (n <= 2) return 0;
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// let compositionCount = 0;
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// for(let i = 3; i < n; i++) {
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// for(let j = i - 1; j > 1 ; j--) {
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// if (i % j === 0) {
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// compositionCount++;
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// break; // 找到一个就可以证明它不是质数了
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// }
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// }
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// }
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// return n - compositionCount - 2; // 需要减去1和n这两个数字
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43+
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// 上面的方法会超时，因此我们需要进行优化
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// 数学角度来看，如果一个数字可以分解为两个数字相乘（这两个数字不包括0和它本身），那么它就是合数
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constcompositions=[];// compositions[i] 表示i是否是合数
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letcount=0;
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for(leti=2;i<n;i++){
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if(!compositions[i])count++;
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for(letj=2;i*j<n;j++){
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compositions[i*j]=true;
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}
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}
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returncount;
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};
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‎backlog/21.merge-two-sorted-lists.js

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/*
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*@lc app=leetcode id=21 lang=javascript
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*
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* [21] Merge Two Sorted Lists
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*
6+
* https://leetcode.com/problems/merge-two-sorted-lists/description/
7+
*
8+
* algorithms
9+
* Easy (46.02%)
10+
* Total Accepted: 562.7K
11+
* Total Submissions: 1.2M
12+
* Testcase Example: '[1,2,4]\n[1,3,4]'
13+
*
14+
* Merge two sorted linked lists and return it as a new list. The new list
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* should be made by splicing together the nodes of the first two lists.
16+
*
17+
* Example:
18+
*
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* Input: 1->2->4, 1->3->4
20+
* Output: 1->1->2->3->4->4
21+
*
22+
*
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*/
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
28+
* this.next = null;
29+
* }
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*/
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/**
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*@param {ListNode} l1
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*@param {ListNode} l2
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*@return {ListNode}
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*/
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varmergeTwoLists=function(l1,l2){
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letcurrent=newListNode();
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constdummy=current;
39+
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while(l1||l2){
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if(!l1){
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current.next=l2;
43+
returndummy.next;
44+
}elseif(!l2){
45+
current.next=l1;
46+
returndummy.next;
47+
}
48+
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if(l1.val<=l2.val){
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current.next=l1;
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l1=l1.next;
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}else{
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current.next=l2;
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l2=l2.next;
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}
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current=current.next;
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}
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returndummy.next;
61+
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// if (l1 === null) return l2;
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// if (l2 === null) return l1;
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// if (l1.val < l2.val) {
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// l1.next = mergeTwoLists(l1.next, l2);
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// return l1;
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// } else {
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// l2.next = mergeTwoLists(l1, l2.next);
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// return l2;
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// }
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};

‎backlog/268.missing-number.js

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/*
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*@lc app=leetcode id=268 lang=javascript
3+
*
4+
* [268] Missing Number
5+
*
6+
* https://leetcode.com/problems/missing-number/description/
7+
*
8+
* algorithms
9+
* Easy (47.60%)
10+
* Total Accepted: 267.7K
11+
* Total Submissions: 556.2K
12+
* Testcase Example: '[3,0,1]'
13+
*
14+
* Given an array containing n distinct numbers taken from 0, 1, 2, ..., n,
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* find the one that is missing from the array.
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*
17+
* Example 1:
18+
*
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*
20+
* Input: [3,0,1]
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* Output: 2
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*
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*
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* Example 2:
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*
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*
27+
* Input: [9,6,4,2,3,5,7,0,1]
28+
* Output: 8
29+
*
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*
31+
* Note:
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* Your algorithm should run in linear runtime complexity. Could you implement
33+
* it using only constant extra space complexity?
34+
*/
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/**
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*@param {number[]} nums
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*@return {number}
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*/
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varmissingNumber=function(nums){
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// 缺失的数字一定是 0 到 n 之间的一个数字
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// 这是一道数论的题目
43+
// 这里用到了一条性质： sum([1,n]) = n * (n+1) / 2
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letsum=0;
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for(letnumofnums)
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sum+=num;
47+
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return(nums.length*(nums.length+1))/2-sum;
49+
};
50+

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