简单思路：循环抽出一个数字作为基数（如果和之前的数组已一样，跳过），然后有两个指针从前往后和从后往前遍历，如果得出的结果小于 0，左边的指针 + 1；如果得出的结果大于 0 ，右边的指针 - 1；相等则 push 进我们的 res 数组，并且一直 left++ 和 right– 直到和上次的数字不同，循环上面的步骤
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/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function (nums) {
nums = nums.sort((a, b) => a - b);
const res = [];
let pre = undefined;
for (let index = 0; index < nums.length - 2; index++) {
const cur = nums[index];
if (cur === pre) {
continue;
}
pre = cur;
let leftIndex = index + 1;
let rightIndex = nums.length - 1;
while (leftIndex < rightIndex) {
let left = nums[leftIndex];
let right = nums[rightIndex];
const sum = cur + left + right;
if (sum < 0) leftIndex += 1;
else if (sum > 0) rightIndex -= 1;
else {
res.push([cur, left, right]);
while (nums[leftIndex + 1] === left && leftIndex < rightIndex)
leftIndex++;
while (nums[rightIndex - 1] === right && leftIndex < rightIndex)
rightIndex--;
leftIndex++;
rightIndex--;
}
}
}
return res;
};
console.log(threeSum([0, 0, 0, 0]));
