feat: 添加三数之和
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leetcode/中等/三数之和.js
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59
leetcode/中等/三数之和.js
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/*
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给定一个包含 n 个整数的数组 nums,判断 nums 中是否存在三个元素 a ,b ,c ,使得 a + b + c = 0 ?请找出所有和为 0 且 不重复 的三元组。
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示例 1:
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输入:nums = [-1,0,1,2,-1,-4]
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输出:[[-1,-1,2],[-1,0,1]]
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示例 2:
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输入:nums = []
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输出:[]
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示例 3:
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输入:nums = [0]
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输出:[]
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提示:
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0 <= nums.length <= 3000
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-105 <= nums[i] <= 105
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*/
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/**
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* @param {number[]} nums
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* @return {number[][]}
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*/
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const threeSum = (nums) => {
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const result = [];
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nums.sort((a, b) => a - b); // 对数组进行排序,方便找取target
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const len = nums.length;
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for (let i = 0; i < len; i++) {
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if (nums[i] > 0) break; // 由于数组是有序的,如果第一个数大于0,后面的无论怎么相加都会大于零
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if (i > 0 && nums[i] === nums[i - 1]) continue; // 出现重复直接跳过本次循环
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let left = i + 1;
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let right = len - 1;
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while (left < right) {
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const sum = nums[i] + nums[left] + nums[right];
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if (sum < 0) {
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left++;
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} else if (sum > 0) {
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right--;
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} else {
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// nums[left] + nums[right] === target, 这就是我们要找的结果集,
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result.push([nums[i], nums[left], nums[right]]);
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// 如果left后面的数与left相等就跳过
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while (left < right && nums[left] === nums[left + 1]) left++;
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// 如果right前面的数与right相等就跳过
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while (left < right && nums[right] === nums[right - 1]) right--;
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left++;
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right--;
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}
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}
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}
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return result;
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};
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console.log(threeSum([-2, 0, 3, -1, 4, 0, 3, 4, 1, 1, 1, -3, -5, 4, 0]));
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