[LeetCode]3Sum

最新推荐文章于 2022-01-27 22:44:32 发布

原创最新推荐文章于 2022-01-27 22:44:32 发布 · 371 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

149 篇文章

订阅专栏

本文介绍了一种高效算法，用于在给定数组中查找所有和为零的唯一三元组。通过先排序数组，再使用双指针技巧，实现O(N^2)的时间复杂度。详细解释了如何避免重复解，并提供了完整代码实现。

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

Note:

Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2 -1 -4},

    A solution set is:
    (-1, 0, 1)
    (-1, -1, 2)

排序然后夹逼，最后复杂度O(N^2).注意重复数字的情况。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> ret;
        sort(nums.begin(),nums.end());
        int len = nums.size();
        const int target = 0;
        for(int i=0;i<len-2;++i){
            int j = i+1;
            int k = len-1;
            if(i>0&&nums[i]==nums[i-1]) continue;//子解
            while(j<k){
                if(nums[i]+nums[j]+nums[k]==target){
                    ret.push_back({ nums[i],nums[j],nums[k] });
                    while(nums[j]==nums[j+1]&&nums[k]==nums[k-1]&&j<k){
                        j++;
                        k--;
                    }
                    j++;
                    k--;
                }
                else if(nums[i]+nums[j]+nums[k]<target){
                    while(nums[j]==nums[j+1]&&j<k) ++j;
                    ++j;
                }
                else if(nums[i]+nums[j]+nums[k]>target){
                    while(nums[k]==nums[k-1]&&j<k) --k;
                    --k;
                }
            }
        }
        return ret;
    }
};