题目描述
给你一个整数数组 nums ，判断是否存在三元组 [nums[i], nums[j], nums[k]] 满足 i != j、i != k 且 j != k ，同时还满足 nums[i] + nums[j] + nums[k] == 0 。请你返回所有和为 0 且不重复的三元组。
注意：答案中不可以包含重复的三元组。
示例 1：
输入：nums = [-1,0,1,2,-1,-4]
输出：[[-1,-1,2],[-1,0,1]]
解释：
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0 。
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0 。
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0 。
不同的三元组是 [-1,0,1] 和 [-1,-1,2] 。
注意，输出的顺序和三元组的顺序并不重要。
示例 2：
输入：nums = [0,1,1]
输出：[]
解释：唯一可能的三元组和不为 0 。
示例 3：
输入：nums = [0,0,0]
输出：[[0,0,0]]
解释：唯一可能的三元组和为 0 。
提示：
3 <= nums.length <= 3000
-10^5 <= nums[i] <= 10^5
题解
快排 + 双指针法
代码
/*** Return an array of arrays of size *returnSize.* The sizes of the arrays are returned as *returnColumnSizes array.* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().*/
// Define a comparison function for qsort to sort integers in ascending order
int cmp(const void *a, const void *b)
{return *(int*)a - *(int*)b;
}// Function to find all unique triplets that sum up to zero
int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes)
{*returnSize = 0; // Initialize the return size to 0if(numsSize < 3)return NULL; // If the input array has less than 3 elements, return NULL as no triplet can be formedqsort(nums, numsSize, sizeof(int), cmp); // Sort the input array in ascending order using the custom comparison function// Allocate memory for the 2D array to store the triplets and their sizesint **ans = (int **)malloc(sizeof(int *) * numsSize * numsSize);*returnColumnSizes = (int *)malloc(sizeof(int) * numsSize * numsSize);int i, j, k, sum;int indexLeft = 0;int indexMiddle = 0;int indexRight = 0;// Use three pointers to traverse the sorted array and find triplets that sum up to zerofor (indexLeft = 0; indexLeft < numsSize - 2; indexLeft++){if (nums[indexLeft] > 0) {// If the current element is greater than 0, no triplet can sum up to zero, return the result/* if nums[indexLeft] is greater than 0, * the function immediately returns ans . * In this case, ans will be returned without any modifications or additional calculations. * The ans variable is a pointer to a 2D array of integers,* and at that point in the code, it would be returned as it is, * without any triplets being calculated or added to it.*/return ans;}// Skip duplicate values/* the keyword continue is used within a loop after a condition is met. * When continue is encountered, * it causes the program flow to immediately jump to the next iteration of the loop * without executing the remaining code within the loop for the current iteration.* This helps in avoiding processing duplicate elements and * ensures that only unique elements are considered during the loop execution. */if (indexLeft > 0 && nums[indexLeft] == nums[indexLeft - 1])continue;indexMiddle = indexLeft + 1;indexRight = numsSize - 1;// Use two pointers to find all possible tripletswhile (indexMiddle < indexRight){sum = nums[indexLeft] + nums[indexMiddle] + nums[indexRight];if (sum == 0){// If the sum is zero, store the triplet in the ans arrayans[*returnSize] = (int*)malloc(sizeof(int) * 3);(*returnColumnSizes)[*returnSize] = 3;ans[*returnSize][0] = nums[indexLeft];ans[*returnSize][1] = nums[indexMiddle];ans[*returnSize][2] = nums[indexRight];*returnSize += 1;// Skip duplicate values// This step is crucial to avoid considering the same combination of elements multiple times and ensures that only unique triplets are recorded. while (indexMiddle < indexRight && nums[indexMiddle] == nums[++indexMiddle]);while (indexMiddle < indexRight && nums[indexRight] == nums[--indexRight]);}else if (sum > 0){// If the sum is greater than zero, decrement the right pointerindexRight--;}else{// If the sum is less than zero, increment the middle pointerindexMiddle++;}}}return ans; // Return the 2D array containing the unique triplets
}
作者：我自横刀向天笑
链接：https://leetcode.cn/problems/3sum/solutions/1211127/san-shu-zhi-he-cyu-yan-xiang-jie-chao-ji-08q2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
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