242.有效的字母异位词

分别定义两个字母哈希表就可以了

class Solution {
public:bool isAnagram(string s, string t) {int ns = s.size(); int ts = t.size();vector<int> vns(26, 0); vector<int> vts(26, 0);if(ns != ts) return false;for(int i = 0;i<ns;i++){vns[s[i] - 'a']++;vts[t[i] - 'a']++;}for(int i = 0;i<26;i++){if(vns[i] != vts[i]) return false;}return true;}
};

349.两个数组的交集

这里关键点是掌握哈希表和vector的相互转化

class Solution {
public:vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {unordered_set<int> numbers(nums1.begin(), nums1.end());unordered_set<int> result_set;for(int i = 0;i<nums2.size();i++){if(numbers.count(nums2[i])) result_set.insert(nums2[i]);}return vector<int>(result_set.begin(), result_set.end());}
};

202.快乐数

这道题的关键是无限循环意味着sum会出现重复的值，这样我们就能提前终止循环

class Solution {
public:int calc(int n){int result = 0;while(n > 0){int num = n % 10;n = n / 10;result += num * num;}return result;}bool isHappy(int n) {unordered_set<int> s;while(n != 1){if(s.count(n)) return false;s.insert(n);n = calc(n);}return true;}
};

1.两数之和

掌握unordered_map的API

class Solution {
public:vector<int> twoSum(vector<int>& nums, int target) {unordered_map<int,int> nums_map;for(int i =0;i<nums.size();i++){nums_map[nums[i]] = i;}for(int i =0;i<nums.size();i++){if(nums_map.count(target - nums[i]) && i != nums_map[target - nums[i]]){return vector<int>({i, nums_map[target - nums[i]]});}}return vector<int>({-1, -1});}
};

常用API