leetcode 15 3sum

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function (nums) {
    const length = nums.length;
    const result = [],
        results = {};
    nums.sort((a, b) => a - b);//先对数组排序
    for (let i = 0; i < length; i++) {
        let left = i + 1, //这里的左端必须要从i+1开始，不仅可以减少比较的次数还能起到去重的作用
            right = length - 1;
        if (i > 0 && nums[i] === nums[i - 1]) {
            continue; //这里是在[1,1,1,1,1]这种情况下只对一个1进行查找
        }
        while (left < right) {
            if (left === i) {
                left++;
                continue;
            }
            if (right === i) {
                right--;
                continue;
            }
            let current = nums[i] + nums[left] + nums[right];
            if (current === 0) {
                let temp = [nums[i], nums[left], nums[right]]
                    result.push(temp)
                        //优化 也是避免了相同数字下的多次比较问题
                    let lefttemp = nums[left],
                        righttemp = nums[right];
                    while (left < right && lefttemp === nums[left]) {
                        left++;
                    }
                    while (left < right && righttemp === nums[right]) {
                        right--;
                    }
                }
                //前面的数组排序就是为了下面这里
                if (current > 0) {
                    right--;//和大于0就右边的减减，这样和才会减少
                }
                if (current < 0) {
                    left++;
                }
            }
        }
        return result;
    };