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LeetCode 刷题【18. 四数之和】

news 2025/8/2 0:02:19

18. 四数之和

自己做

解：三重循环+双指针

class Solution {
public:vector<vector<int>> fourSum(vector<int>& nums, int target) {int len = nums.size();vector<vector<int>> res;sort(nums.begin(), nums.end());                      //双指针前的排序for (int a = 0; a < len - 3; a++) {while (a > 0 && a < len - 3 && nums[a] == nums[a - 1])           //a不重复【a移动后发现和原来的值相比并没有变化】a++;for (int b = a + 1; b < len - 2; b++) {while (b > a + 1 && b < len - 2 && nums[b] == nums[b - 1])           //b不重复b++;int c = b + 1;int d = len - 1;while (c < d) {//int类型的范围是（有符号）-2**31 ~ 2**31 - 1，即2*10**9左右，（一个数的上限是1*10**9，即至少前两个数相加不可能溢出）对于溢出的情况处理：int x = nums[a] + nums[b];int y = nums[c] + nums[d];if((x > 0 && y > 0 && y > std::numeric_limits<int>::max() - x) ||                    //都为正的情况下相加溢出(x < 0 && y < 0 && y < std::numeric_limits<int>::min() - x) ){                   //都为负的情况下相加溢出//溢出说明组合肯定不成立，开始找下一个组合//(这里移动c、d谁都行)c++;}else{                               //不会溢出的情况if (x + y == target) {                         //相等存放进结果里res.push_back(vector<int>({ nums[a] , nums[b] , nums[c] , nums[d] }));//(这里移动c、d谁都行，但是要保证组合有变化)c++;while (c < d && nums[c] == nums[c - 1]) {                                          //c不重复c++;}}if (x + y > target) {                         //d大了d--;}if (x + y < target) {                         //c小了c++;}}}}}return res;}
};

看题解

优化自己的代码

优化不了一点

官方代码

class Solution {
public:vector<vector<int>> fourSum(vector<int>& nums, int target) {vector<vector<int>> quadruplets;if (nums.size() < 4) {return quadruplets;}sort(nums.begin(), nums.end());int length = nums.size();for (int i = 0; i < length - 3; i++) {if (i > 0 && nums[i] == nums[i - 1]) {continue;}if ((long) nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {break;}if ((long) nums[i] + nums[length - 3] + nums[length - 2] + nums[length - 1] < target) {continue;}for (int j = i + 1; j < length - 2; j++) {if (j > i + 1 && nums[j] == nums[j - 1]) {continue;}if ((long) nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {break;}if ((long) nums[i] + nums[j] + nums[length - 2] + nums[length - 1] < target) {continue;}int left = j + 1, right = length - 1;while (left < right) {long sum = (long) nums[i] + nums[j] + nums[left] + nums[right];if (sum == target) {quadruplets.push_back({nums[i], nums[j], nums[left], nums[right]});while (left < right && nums[left] == nums[left + 1]) {left++;}left++;while (left < right && nums[right] == nums[right - 1]) {right--;}right--;} else if (sum < target) {left++;} else {right--;}}}}return quadruplets;}
};