[LeetCode] 15. 3Sum

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

三数之和。题意是给一个input数组，请问是否能找到这样的组合使得a + b + c = 0？输出所有这样的[a, b, c]组合。

还是two sum的思路，双指针从两边往中间逼近。因为和是0，所以可以先固定a的位置，然后找是否有b + c = 0 - a的组合。首先需要对数组排序，然后在执行过程中需要排除一些重复数字，避免最后的output有重复。同时，在遍历a的范围的时候，如果a > 0就可以直接跳过了，因为数组是排序过的，如果a大于0，b，c也必定大于0。

时间O(n^2)

空间O(n^2) - list of list

JavaScript实现

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(nums) {
    const res = [];
    let left, right, sum;
    nums.sort((a, b) => a - b);
    for (let i = 0; i < nums.length - 2; i++) {
        if (nums[i] > 0) break;
        if (nums[i] === nums[i - 1]) continue;
        left = i + 1;
        right = nums.length - 1;
        sum = 0 - nums[i];
        while (left < right) {
            if (nums[left] + nums[right] === sum) {
                res.push([nums[i], nums[left], nums[right]]);
                while (left < right && nums[left] === nums[left + 1]) {
                    left++;
                }
                while (left < right && nums[right] === nums[right - 1]) {
                    right--;
                }
                left++;
                right--;
            } else if (nums[left] + nums[right] < sum) {
                left++;
            } else {
                right--;
            }
        }
    }
    return res;
};

Java实现

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int sum = 0 - nums[i];
            int low = i + 1;
            int high = nums.length - 1;
            while (low < high) {
                if (nums[low] + nums[high] == sum) {
                    res.add(Arrays.asList(nums[i], nums[low], nums[high]));
                    while (low < high && nums[low] == nums[low + 1]) {
                        low++;
                    }
                    while (low < high && nums[high] == nums[high - 1]) {
                        high--;
                    }
                    low++;
                    high--;
                } else if (nums[low] + nums[high] < sum) {
                    low++;
                } else {
                    high--;
                }
            }
        }
        return res;
    }
}

two sum题目总结

LeetCode 题目总结