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Commitad9f834

Browse files
author
zhenyu zhang
committed
update LeetCode 18、34、40
1 parentc1d5321 commitad9f834

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6 files changed

+254
-2
lines changed

6 files changed

+254
-2
lines changed

‎src/com/hadley/Main.java

Lines changed: 28 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,34 @@
11
packagecom.hadley;
22

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importjava.util.Stack;
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36
publicclassMain {
47

5-
publicstaticvoidmain(String[]args) {
6-
// write your code here
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publicintlongestValidParentheses(Strings) {
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intresult =0;
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//用栈来进行处理
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Stack<Integer>stack =newStack<>();
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intindex = -1;//指针
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for(inti =0;i <s.length();i++){
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if(s.charAt(i) =='('){
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stack.push(i);
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}else{//s.charAt(i) == ')'
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if(stack.isEmpty()){
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index =i;
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}else{
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stack.pop();
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if(stack.isEmpty()){//i - index是当前的子串长度
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result =Math.max(result,i -index);
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}else{
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result =Math.max(result,i -stack.peek());
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}
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}
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}
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}
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returnresult;
732
}
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}

‎src/com/hadley/Main1.java

Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
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packagecom.hadley;
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importjava.util.Arrays;
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importjava.util.Scanner;
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publicclassMain1 {
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publicstaticvoidmain(String[]args) {
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Scannersc =newScanner(System.in);
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intpeople =sc.nextInt();
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inttime =sc.nextInt();
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int[]coinList =newint[people];
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for(inti =0;i <people;i++){
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coinList[i] =sc.nextInt();
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}
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Arrays.sort(coinList);
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intmax =coinList[people-1];
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intmin =coinList[0];
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}
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}

‎src/com/hadley/_018.java

Lines changed: 42 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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packagecom.hadley;
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/*
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2020.05.24 18.4Sum
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Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
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Note:
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The solution set must not contain duplicate quadruplets.
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Example:
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Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.
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A solution set is:
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[
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[-1, 0, 0, 1],
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[-2, -1, 1, 2],
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[-2, 0, 0, 2]
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]
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*/
22+
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importjava.util.ArrayList;
24+
importjava.util.Arrays;
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importjava.util.List;
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publicclass_018 {
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publicList<List<Integer>>fourSum(int[]nums,inttarget) {
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Arrays.sort(nums);
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List<List<Integer>>resultList =newArrayList<>();
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//the difference between ArrayList and List：https://www.cnblogs.com/zcscnn/p/7743507.html
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33+
if(nums.length <4){
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returnresultList;
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}else{
36+
//TODO
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returnnull;
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}
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40+
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}
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}

‎src/com/hadley/_034.java

Lines changed: 83 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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packagecom.hadley;
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/*
4+
2020.06.20
5+
Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.
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Your algorithm's runtime complexity must be in the order of O(log n).
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If the target is not found in the array, return [-1, -1].
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Example 1:
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Input: nums = [5,7,7,8,8,10], target = 8
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Output: [3,4]
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Example 2:
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Input: nums = [5,7,7,8,8,10], target = 6
18+
Output: [-1,-1]
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*/
20+
21+
publicclass_034 {
22+
23+
//time complexity: O(n)
24+
publicint[]searchRange(int[]nums,inttarget) {
25+
int[]result = {-1,-1};
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booleanflag =false;
28+
29+
for(inti =0;i <nums.length;i++){
30+
if(nums[i] <target){
31+
continue;
32+
}
33+
34+
if(nums[i] ==target && !flag){
35+
result[0] =i;
36+
result[1] =i;
37+
flag =true;
38+
}elseif(nums[i] ==target &&flag){
39+
result[1] =i;
40+
}
41+
}
42+
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returnresult;
44+
}
45+
46+
//come from the Solution time complexity: O(logN)
47+
publicint[]searchRange1(int[]nums,inttarget){
48+
intleft =0;
49+
intright =nums.length -1;
50+
intl =left;
51+
intr =right;
52+
53+
//find the left border of target
54+
while(left <right){
55+
intmid = (left +right)/2;
56+
if(nums[mid] <target){
57+
left =mid +1;
58+
}else{
59+
right =mid;
60+
}
61+
}
62+
63+
if(left >right||nums[left]!=target){
64+
returnnewint[]{-1,-1};
65+
}
66+
67+
//find the right border of target
68+
while(l <r){
69+
intmid = (l +r)/2 +1;
70+
if(nums[mid] >target){
71+
r =mid -1;
72+
}else{
73+
l =mid;
74+
}
75+
}
76+
77+
if(left >right ||left >r){
78+
returnnewint[]{-1,-1};
79+
}else{
80+
returnnewint[]{left,r};
81+
}
82+
}
83+
}

‎src/com/hadley/_040.java

Lines changed: 71 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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packagecom.hadley;
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importjava.util.ArrayList;
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importjava.util.Arrays;
5+
importjava.util.List;
6+
7+
/*
8+
2020.05.07
9+
Given a collection of candidate numbers (candidates) and a target number (target),
10+
find all unique combinations in candidates where the candidate numbers sums to target.
11+
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Each number in candidates may only be used once in the combination.
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Note:
15+
16+
All numbers (including target) will be positive integers.
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The solution set must not contain duplicate combinations.
18+
Example 1:
19+
20+
Input: candidates = [10,1,2,7,6,1,5], target = 8,
21+
A solution set is:
22+
[
23+
[1, 7],
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[1, 2, 5],
25+
[2, 6],
26+
[1, 1, 6]
27+
]
28+
Example 2:
29+
30+
Input: candidates = [2,5,2,1,2], target = 5,
31+
A solution set is:
32+
[
33+
[1,2,2],
34+
[5]
35+
]
36+
*/
37+
publicclass_040 {
38+
39+
publicList<List<Integer>>combinationSum2(int[]candidates,inttarget) {
40+
List<List<Integer>>ans =newArrayList<>();
41+
Arrays.sort(candidates);
42+
dfs(candidates,target,0,newArrayList<Integer>(),ans);
43+
44+
returnans;
45+
}
46+
47+
/*
48+
curList = current operating List
49+
*/
50+
publicvoiddfs(int[]candidates,inttarget,intindex,List<Integer>curList,List<List<Integer>>ans){
51+
if(target ==0){
52+
ans.add(newArrayList<>(curList));
53+
return;
54+
}
55+
56+
for(inti =index;i <candidates.length;i++){
57+
if(candidates[i] >target){
58+
break;
59+
}
60+
61+
//skip the same element
62+
if(i >index &&candidates[i] ==candidates[i-1]){
63+
continue;
64+
}
65+
curList.add(candidates[i]);
66+
dfs(candidates,target -candidates[i],i +1,curList,ans);
67+
curList.remove(curList.size()-1);
68+
}
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}
71+
}
Lines changed: 6 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,6 @@
1+
packagecom.hadley.challenge30days;
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3+
publicclassLRUCache {
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5+
//TODO
6+
}

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