Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.
/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function(nums) {
if (nums.length < 3) return [];
nums.sort((a, b) => a - b);
const ans = [];
for (let i = 0; i < nums.length - 2; i++) {
if (nums[i] > 0) break;
if (i > 0 && nums[i] === nums[i - 1]) continue;
let j = i + 1,
k = nums.length - 1;
while (j < k) {
const sum = nums[i] + nums[j] + nums[k];
if (sum === 0) {
ans.push([nums[i], nums[j], nums[k]]);
while (nums[j] === nums[j + 1]) j++;
while (nums[k] === nums[k - 1]) k--;
j++;
k--;
} else if (sum > 0) k--;
else j++;
}
}
return ans;
};
- 排序完後只需控另外兩個，做法類似 167.
- 要注意的是不能出現重複組合，j、k 在 while 中處理但 i 要回頭方式 check，不然遇到
[ -1, -1, 0, 2]這種就把需要的跳過了