1--有效的字母异位词

利用哈希表存储每个字母的出现次数，比较两个字符串各个字母出现次数是否相等即可；

#include <iostream>
#include <string>
#include <vector>class Solution {
public:bool isAnagram(std::string s, std::string t) {// 用数组实现哈希表std::vector<int> hash(26, 0);for(auto i = 0; i < s.length(); i++){hash[s[i] - 'a']++; }// 遍历字符串tfor(auto i = 0; i < t.length(); i++){hash[t[i] - 'a']--; }// 遍历哈希表for(int i = 0; i < 26; i++){if(hash[i] != 0) return false;}return true;}
};int main(int argc, char argv[]){// s = "anagram", t = "nagaram"std::string s = "anagram", t = "nagaram";Solution S1;bool res = S1.isAnagram(s, t);if(res) std::cout << "true" << std::endl;else std::cout << "false" << std::endl;return 0;
}

2--两个数组的交集

利用哈希表存储数组1的元素，接着遍历数组2并判断元素是否存储在哈希表中，将交集元素存储并返回；

#include <iostream>
#include <unordered_set>
#include <vector>class Solution {
public:std::vector<int> intersection(std::vector<int>& nums1, std::vector<int>& nums2) {std::unordered_set<int> hash1;for(int i = 0; i < nums1.size(); i++){hash1.insert(nums1[i]);}std::unordered_set<int> result;for(int i = 0; i < nums2.size(); i++){if(hash1.find(nums2[i]) != hash1.end()){result.insert(nums2[i]);}}std::vector<int> res;for(auto item : result){res.push_back(item);}return res;}
};int main(int argc, char argv[]){// nums1 = [1,2,2,1], nums2 = [2,2]std::vector<int> nums1 = {1, 2, 2, 1}, nums2 = {2, 2};Solution S1;std::vector<int> res = S1.intersection(nums1, nums2);for(auto item : res) std::cout << item << " ";std::cout << std::endl;return 0;
}

3--两数之和

利用哈希表存储，key 为 target - nums[i]，value 为 i；遍历数组判断当前nums[i]是否在哈希表中出现，返回匹配的两个结果对应的索引即可；

#include <iostream>
#include <unordered_map>
#include <vector>class Solution {
public:std::vector<int> twoSum(std::vector<int>& nums, int target) {std::vector<int> res;std::unordered_map<int, int> hash;for(int i = 0; i < nums.size(); i++){if(hash.find(nums[i]) != hash.end()){res.push_back(hash[nums[i]]);res.push_back(i);return res;}hash.emplace(target - nums[i], i);}return res;}
};int main(int argc, char argv[]){// nums = [2,7,11,15], target = 9std::vector<int> nums = {2, 7, 11, 15};int target = 9;Solution S1;std::vector<int> res = S1.twoSum(nums, target);for(auto item : res) std::cout << item << " ";std::cout << std::endl;return 0;
}

4--四数相加II

利用哈希表，其中key为两个数组对应元素的和，value为出现的次数；接着遍历剩下两个数组的元素和是否在哈希表中出现，记录出现的次数即匹配结果；

#include <iostream>
#include <unordered_map>
#include <vector>class Solution {
public:int fourSumCount(std::vector<int>& nums1, std::vector<int>& nums2, std::vector<int>& nums3, std::vector<int>& nums4) {std::unordered_map<int, int> hash;for(int i = 0; i < nums1.size(); i++){for(int j = 0; j < nums2.size(); j++){hash[(nums1[i] + nums2[j])] ++;}}int res = 0;for(int i = 0; i < nums3.size(); i++){for(int j = 0; j < nums3.size(); j++){if(hash.find(0 - (nums3[i] + nums4[j])) != hash.end()){res += hash[0 - (nums3[i] + nums4[j])];}}}return res;}
};int main(int argc, char argv[]){// nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]std::vector<int> nums1 = {1, 2}, nums2 = {-2, -1}, nums3 = {-1, 2}, nums4 = {0, 2};Solution S1;int res = S1.fourSumCount(nums1, nums2, nums3, nums4);std::cout << res << std::endl;return 0;
}

5--三数之和

本题基于双指针算法，需要注意去重；

#include <iostream>
#include <vector>
#include <algorithm>class Solution {
public:std::vector<std::vector<int>> threeSum(std::vector<int>& nums) {std::vector<std::vector<int>> res;std::sort(nums.begin(), nums.end());for(int i = 0; i < nums.size(); i++){if(i > 0 && nums[i] == nums[i - 1]) continue; // 去重int l = i + 1, r = nums.size() - 1;while(l < r){if(l > i + 1 && nums[l] == nums[l - 1]){l++;continue; // 去重}if(nums[i] + nums[l] + nums[r] == 0){res.push_back({nums[i], nums[l], nums[r]});l++;r--;}else if(nums[i] + nums[l] + nums[r] > 0){r--;}else{l++;}}}return res;}
};int main(int argc, char* argv[]){// nums = [-1,0,1,2,-1,-4]std::vector<int> nums = {-1, 0, 1, 2, -1, -4};Solution S1;std::vector<std::vector<int>> res = S1.threeSum(nums);for(auto item : res){for(auto v : item) std::cout << v << " ";std::cout << std::endl;}return 0;
}

6--四数之和

类似于三数之和，只需额外多遍历一次，同时注意剪枝和去重的操作；

#include <iostream>
#include <vector>
#include <algorithm>class Solution {
public:std::vector<std::vector<int>> fourSum(std::vector<int>& nums, int target) {std::vector<std::vector<int>> res;std::sort(nums.begin(), nums.end());int len = nums.size();for(int i = 0; i < len; i++){// 剪枝if(nums[i] >= 0 && nums[i] > target) break;// 去重if(i > 0 && nums[i] == nums[i-1]) continue;// three sumfor(int j = i + 1; j < len; j++){// 剪枝if (nums[i] + nums[j] > target && nums[i] + nums[j] >= 0) break;// 去重if (j > i + 1 && nums[j] == nums[j - 1]) continue;int l = j + 1, r = len - 1;while(l < r){if((long) nums[i] + nums[j] + nums[l] + nums[r] == target){res.push_back({nums[i], nums[j], nums[l], nums[r]});// 去重while (r > l && nums[r] == nums[r - 1]) r--;while (r > l && nums[l] == nums[l + 1]) l++;l++;r--;}else if((long) nums[i] + nums[j] + nums[l] + nums[r] < target){l++;}else{r--;}}}}return res;}
};int main(int argc, char* argv[]){// nums = [-1,0,1,2,-1,-4]std::vector<int> nums = {1, 0, -1, 0, -2, 2};int target = 0;Solution S1;std::vector<std::vector<int>> res = S1.fourSum(nums, target);for(auto item : res){for(auto v : item) std::cout << v << " ";std::cout << std::endl;}return 0;
}