思路就是先统计nums1和num2各个元素之和出现的次数，然后遍历num3和nums4各个元素之和，看其相反数是否在map中，若在加上出现次数

class Solution {
public:
int fourSumCount(vector<int> &nums1, vector<int> &nums2, vector<int> &nums3, vector<int> &nums4) {unordered_map<int, int> map;for (auto &ele1 : nums1) {for (auto &ele2 : nums2) {map[ele1 + ele2]++;}}int res = 0;for (auto &ele1 : nums3) {for (auto &ele2 : nums4) {auto it = map.find(-(ele1 + ele2));if (it != map.end()) {res += it->second;}}}return res;
}
};

感觉和前面某题很类似，用map和数组都可以实现

class Solution {
public:bool canConstruct(string ransomNote, string magazine) {vector<int> arr(26, 0);for (int i = 0; i < magazine.size(); i++) {arr[magazine[i] - 'a']++;}for (int i = 0; i < ransomNote.size(); i++) {arr[ransomNote[i] - 'a']--;if (arr[ransomNote[i] - 'a'] < 0) {return false;}}return true;
}
};class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {unordered_map<char, int> map;for (int i = 0; i < magazine.size(); i++) {map[magazine[i]]++;}for (int i = 0; i < ransomNote.size(); i++) {map[ransomNote[i]]--;if (map[ransomNote[i]] < 0) return false;}return true;
}
};

卡哥的视频讲得很清楚

class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {vector<vector<int>> res;sort(nums.begin(), nums.end());for (int i = 0; i < nums.size(); i++) {if (nums[i] > 0) continue;if (i > 0 && nums[i] == nums[i - 1]) continue;int left = i + 1;int right = nums.size() - 1;while (left < right) {int sum = nums[i] + nums[left] + nums[right];if (sum < 0) {left++;} else if (sum > 0) {right--;} else {res.push_back({nums[i], nums[left], nums[right]});while (left < right && nums[left] == nums[left + 1]) left++;while (left < right && nums[right] == nums[right - 1]) right--;left++;right--;}}}return res;
}
};

在三数之和外面套一层for

需要注意的点是这里的target可以是负数所以在第一层剪枝除了要>target 还要>=0

其次会发现和越界[0,0,0,1000000000,1000000000,1000000000,1000000000]，1000000000 + 1000000000 + 1000000000就会越界所以我才用了long long类型为和，或者只加前面三个，判断target-nums[right]

class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {sort(nums.begin(), nums.end());vector<vector<int>> res;for (int i = 0; i < nums.size(); i++) {if (nums[i] > target && nums[i] >= 0) continue;if (i > 0 && nums[i] == nums[i - 1]) continue;for (int j = i + 1; j < nums.size(); j++) {if (nums[i] + nums[j] > target && nums[i] + nums[j] >= 0) continue;if (j > i + 1 && nums[j] == nums[j - 1]) continue;int left = j + 1;int right = nums.size() - 1;while (left < right) {long long sum = (long long)0 + nums[i] + nums[j] + nums[left] + nums[right];if (sum < target) {left++;} else if (sum > target) {right--;} else {res.push_back({nums[i], nums[j], nums[left], nums[right]});while (left < right && nums[left] == nums[left + 1]) left++;while (left < right && nums[right] == nums[right - 1]) right--;left++;right--;}}}}return res;
}
};