【力扣】18. 四数之和

18. 四数之和

题目描述

给你一个由 n 个整数组成的数组 nums ，和一个目标值 target 。请你找出并返回满足下述全部条件且不重复的四元组 [nums[a], nums[b], nums[c], nums[d] （若两个四元组元素一一对应，则认为两个四元组重复）：

0 <= a, b, c, d < n
a、b、c 和 d 互不相同
nums[a] + nums[b] + nums[c] + nums[d] == target
你可以按任意顺序返回答案。

示例 1：

输入：nums = [1,0,-1,0,-2,2], target = 0
输出：[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]

示例 2：

输入：nums = [2,2,2,2,2], target = 8
输出：[[2,2,2,2]]

提示：

1 <= nums.length <= 200
-10^9^ <= nums[i] <= 10^9^
-10^9^ <= target <= 10^9^

解题方法

排序+双指针

C 语言

c 复制代码

/**
 * 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().
 */
int my_cmp(const void* a, const void* b) { return *(int*)a - *(int*)b; }

int** fourSum(int* nums, int numsSize, int target, int* returnSize,
              int** returnColumnSizes) {

    int** quadruplets = malloc(sizeof(int*) * 1001);
    *returnColumnSizes = malloc(sizeof(int) * 1001);
    *returnSize = 0;

    if (numsSize < 4) {
        return quadruplets;
    }

    qsort(nums, numsSize, sizeof(int), my_cmp);

    for (int i = 0; i < numsSize - 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[numsSize - 3] + nums[numsSize - 2] +
                nums[numsSize - 1] < target) {
            continue;
        }

        for (int j = i + 1; j < numsSize - 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[numsSize - 2] +
                    nums[numsSize - 1] < target) {
                continue;
            }

            int left = j + 1, right = numsSize - 1;

            while (left < right) {

                long sum = (long)nums[i] + nums[j] + nums[left] + nums[right];

                if (sum == target) {
                    int* tmp = malloc(sizeof(int) * 4);
                    tmp[0] = nums[i];
                    tmp[1] = nums[j];
                    tmp[2] = nums[left];
                    tmp[3] = nums[right];
                    (*returnColumnSizes)[(*returnSize)] = 4;
                    quadruplets[(*returnSize)++] = tmp;

                    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;
}