Leetcode题目分类整理

一、数组

8） 双指针 ---- 滑动窗口

例题：

    3. Longest Substring Without Repeating Characters

描述：Given a string, find the length of the longest substring without repeating characters.

题解：时间：92.67%，空间：87.02%

 1     public int lengthOfLongestSubstring(String s) {   
 2         // check
 3         if(s == null || s.length() == 0) return 0;
 4         
 5         // initial 
 6         int[] freq = new int[256];
 7         Arrays.fill(freq, -1);
 8         int l = 0, r = 0, res = 0;
 9         
10         while(r < s.length()){
11             if(freq[s.charAt(r)] != -1) l = Math.max(l, freq[s.charAt(r)] + 1);
12             freq[s.charAt(r)] = r++;
13             res = Math.max(res, r - l);
14         }
15         
16         return res;
17     }

练习：

438. Find All Anagrams in a String

 76. Minimum Window Substring

二、查找问题 

1）查找有无：是否存在

2）查找对应关系：出现了几次

349. Intersection of Two Arrays   // 熟悉Set

描述：Given two arrays, write a function to compute their intersection.

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]

 1     public int[] intersection(int[] nums1, int[] nums2) {
 2         if(nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0) 
 3             return new int[0];
 4         
 5         Set<Integer> set1 = new HashSet<Integer>();
 6         Set<Integer> setTemp = new HashSet<Integer>();
 7         
 8         for(int num : nums1) set1.add(num);
 9         for(int num : nums2) if(set1.contains(num)) setTemp.add(num);
10             
11         int k = setTemp.size();
12         int[] res = new int[k];
13         for(int num : setTemp) res[--k] = num;
14         
15         return res;    
16     }

350. Intersection of Two Arrays II          // 熟悉 Map

描述：Given two arrays, write a function to compute their intersection.

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]

 1     public int[] intersect(int[] nums1, int[] nums2) {
 2         if(nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0)
 3             return new int[0];
 4         
 5         Map<Integer, Integer> map = new HashMap<Integer, Integer>();
 6         for(int num : nums1){
 7             map.put(num, map.getOrDefault(num, 0) + 1);
 8         }
 9         
10         List<Integer> list = new ArrayList<Integer>();
11         for(int num : nums2){
12             if(map.get(num) != null && map.get(num) > 0){
13                 list.add(num);
14                 map.put(num, map.get(num) - 1);
15             }
16         }
17         
18         int k = list.size();
19         int[] arr = new int[k];
20         for(int num : list) arr[--k] = num;
21         
22         return arr;
23     }

242. Valid Anagram

题目描述：Given two strings s and t , write a function to determine if t is an anagram of s.

202. Happy Number

290. Word Pattern

205. Isomorphic Strings

451. Sort Characters By Frequency

查找表的经典问题：

1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

时间：99.77%，空间：88.15% 

1     public int[] twoSum(int[] nums, int target) {
2         Map<Integer, Integer> map = new HashMap<Integer, Integer>();
3         for(int i = 0; i < nums.length; i++){
4             int de = target - nums[i];
5             if(map.get(de) != null) return new int[]{map.get(de), i};
6             map.put(nums[i], i);
7         }
8         return new int[2];
9     }

15. 3Sum

18. 4Sum

16. 3Sum Closest

454. 4Sum II

描述：Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -2^28 to 2^28 - 1 and the result is guaranteed to be at most 2^31 - 1.

49. Group Anagrams

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来源： https://www.cnblogs.com/tf-Y/p/10412101.html