算法-四数之和

leetcode-四数之和
https://leetcode-cn.com/problems/4sum/
坚持每天进步

四数之和

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]] ：

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
• -109 <= nums[i] <= 109
• -109 <= target <= 109

解题思路

• 定义双指针，避免暴力循环解题；

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[][]}
 */
var fourSum = function (nums, target) {

    nums = nums.sort((a, b) => (a - b));
    let res = [];
    let len = nums.length;
    for (let i = 0; i < len - 2; i++) {
       
        if (i > 0 && nums[i] == nums[i - 1]) {
            continue;
        }

        for (let j = i + 1; j < len - 2; j++) {
           
            if (j > i + 1 && nums[j] == nums[j - 1]) {
                continue;
            }

            let firstNum = nums[i];
            let secondNum = nums[j];

            const restVal = target - firstNum - secondNum

            let leftIndex = j + 1;
            let rightIndex = len - 1;
            while(leftIndex < rightIndex) {
                let sum = nums[leftIndex] + nums[rightIndex];
                if (restVal === sum) {

                    res.push([firstNum, secondNum, nums[leftIndex], nums[rightIndex]]);

                    while (nums[leftIndex] === nums[leftIndex + 1]) {
                        leftIndex++;
                    }
                    while (nums[rightIndex] === nums[rightIndex - 1]) {
                        rightIndex--;
                    }
                    leftIndex++;
                    rightIndex--;
                } else if (sum < restVal) {
                    leftIndex++;
                } else if (sum > restVal) {
                    rightIndex--;
                }
            }
        }
    }

    return res;

};