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Commit24e98f4

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1 parent1d9fc2b commit24e98f4

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‎100. Same Tree.go

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This file was deleted.

‎100. [E]Same Tree.go

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package main
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/**
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Given two binary trees, write a function to check if they are the same or not.
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Two binary trees are considered the same if they are structurally identical and the nodes have the same value.
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Example 1:
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Input: 1 1
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/ \ / \
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2 3 2 3
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[1,2,3], [1,2,3]
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Output: true
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Example 2:
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Input: 1 1
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/ \
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2 2
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[1,2], [1,null,2]
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Output: false
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Example 3:
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Input: 1 1
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/ \ / \
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2 1 1 2
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[1,2,1], [1,1,2]
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Output: false
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*/
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typeTreeNodestruct {
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Valint
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Left*TreeNode
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Right*TreeNode
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}
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funcisSameTree(p*TreeNode,q*TreeNode)bool {
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ifp==nil&&q==nil {
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returntrue
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}
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if (p==nil&&q!=nil)|| (p!=nil&&q==nil) {
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returnfalse
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}
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ifp.Val!=q.Val {
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returnfalse
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}
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returnisSameTree(p.Left,q.Left)&&isSameTree(p.Right,q.Right)
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}

‎101. Symmetric Tree.gorenamed to‎101. [E]Symmetric Tree.go

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package main
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/**
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Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
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For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
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1
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/ \
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2 2
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/ \ / \
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3 4 4 3
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But the following [1,2,2,null,3,null,3] is not:
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1
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/ \
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2 2
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\ \
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3 3
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*/
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typeTreeNodestruct {
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Valint
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Left*TreeNode

‎105. Construct Binary Tree from Preorder and Inorder Traversal.gorenamed to‎105. [M]Construct Binary Tree from Preorder and Inorder Traversal.go

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package main
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/**
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Given preorder and inorder traversal of a tree, construct the binary tree.
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Note:
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You may assume that duplicates do not exist in the tree.
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For example, given
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preorder = [3,9,20,15,7]
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inorder = [9,3,15,20,7]
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Return the following binary tree:
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3
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/ \
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9 20
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/ \
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15 7
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*/
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typeTreeNodestruct {
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Valint
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Left*TreeNode

‎78. Subsets.gorenamed to‎78. [M]Subsets.go

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package main
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/**
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Given a set of distinct integers, nums, return all possible subsets (the power set).
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Note: The solution set must not contain duplicate subsets.
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Example:
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Input: nums = [1,2,3]
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Output:
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[
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[3],
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[1],
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[2],
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[1,2,3],
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[1,3],
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[2,3],
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[1,2],
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[]
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]
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*/
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/**
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combination:
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dfs方法，注意中间结果一定要copy到新的变量中，不然会得到空值

‎79. Word Search.gorenamed to‎79. [M]Word Search.go

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package main
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/**
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Given a 2D board and a word, find if the word exists in the grid.
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The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.
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Example:
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board =
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[
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['A','B','C','E'],
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['S','F','C','S'],
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['A','D','E','E']
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]
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Given word = "ABCCED", return true.
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Given word = "SEE", return true.
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Given word = "ABCB", return false.
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*/
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funchelp(b [][]byte,wordstring,i,jint)bool {
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ifword=="" {
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returntrue

‎81. Search in Rotated Sorted Array II.gorenamed to‎81. [M]Search in Rotated Sorted Array II.go

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package main
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/**
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Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
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(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).
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You are given a target value to search. If found in the array return true, otherwise return false.
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Example 1:
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Input: nums = [2,5,6,0,0,1,2], target = 0
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Output: true
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Example 2:
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Input: nums = [2,5,6,0,0,1,2], target = 3
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Output: false
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*/
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funcsearch(nums []int,targetint)bool {
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l:=len(nums)
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ifl==0 {
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package main
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/**
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Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
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Example 1:
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Input: 1->2->3->3->4->4->5
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Output: 1->2->5
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Example 2:
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Input: 1->1->1->2->3
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Output: 2->3
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*/
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typeListNodestruct {
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Valint
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Next*ListNode
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}
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funcdeleteDuplicates(head*ListNode)*ListNode {
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t:=&ListNode{}
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t.Next=head
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p:=t
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forp.Next!=nil&&p.Next.Next!=nil {
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ifp.Next.Val==p.Next.Next.Val {
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dup:=p.Next.Val
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forp.Next!=nil&&p.Next.Val==dup {
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p.Next=p.Next.Next
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}
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}else {
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p=p.Next
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}
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}
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returnt.Next
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}

‎83. Remove Duplicates from Sorted List.gorenamed to‎83. [E]Remove Duplicates from Sorted List.go

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package main
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/**
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Given a sorted linked list, delete all duplicates such that each element appear only once.
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Example 1:
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Input: 1->1->2
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Output: 1->2
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Example 2:
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Input: 1->1->2->3->3
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Output: 1->2->3
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*/
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typeListNodestruct {
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Valint
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Next*ListNode

‎86. Partition List.gorenamed to‎86. [M]Partition List.go

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package main
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/**
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Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
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You should preserve the original relative order of the nodes in each of the two partitions.
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Example:
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Input: head = 1->4->3->2->5->2, x = 3
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Output: 1->2->2->4->3->5
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*/
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typeListNodestruct {
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Valint
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Next*ListNode

‎88. Merge Sorted Array.go

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‎88. [E]Merge Sorted Array.go

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package main
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/**
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Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
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Note:
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The number of elements initialized in nums1 and nums2 are m and n respectively.
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You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.
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Example:
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Input:
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nums1 = [1,2,3,0,0,0], m = 3
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nums2 = [2,5,6], n = 3
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Output: [1,2,2,3,5,6]
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*/
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funcmerge(nums1 []int,mint,nums2 []int,nint) {
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form>0||n>0 {
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ifn==0 {
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break
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}
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ifm==0 {
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nums1[n-1]=nums2[n-1]
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n--
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continue
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}
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ifnums1[m-1]>nums2[n-1] {
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nums1[m+n-1]=nums1[m-1]
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m--
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}else {
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nums1[m+n-1]=nums2[n-1]
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n--
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}
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}
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}

‎89. [M]Gray Code.go

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package main
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/**
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The gray code is a binary numeral system where two successive values differ in only one bit.
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Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.
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Example 1:
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Input: 2
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Output: [0,1,3,2]
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Explanation:
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00 - 0
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01 - 1
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11 - 3
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10 - 2
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For a given n, a gray code sequence may not be uniquely defined.
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For example, [0,2,3,1] is also a valid gray code sequence.
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00 - 0
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10 - 2
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11 - 3
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01 - 1
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Example 2:
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Input: 0
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Output: [0]
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Explanation: We define the gray code sequence to begin with 0.
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A gray code sequence of n has size = 2n, which for n = 0 the size is 20 = 1.
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Therefore, for n = 0 the gray code sequence is [0].
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*/
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funcgrayCode(nint) []int {
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res:= []int{0}
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fori:=0;i<n;i++ {
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h:=1<<uint(i)
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size:=len(res)
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forj:=size-1;j>=0;j-- {
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num:=res[j]
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num+=h
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res=append(res,num)
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}
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}
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returnres
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}

‎92. Reverse Linked List II.gorenamed to‎92. [M]Reverse Linked List II.go

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package main
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/**
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Reverse a linked list from position m to n. Do it in one-pass.
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Note: 1 ≤ m ≤ n ≤ length of list.
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Example:
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Input: 1->2->3->4->5->NULL, m = 2, n = 4
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Output: 1->4->3->2->5->NULL
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*/
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typeListNodestruct {
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Valint
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Next*ListNode

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