leetcode-3sum(n^2)

最新推荐文章于 2021-12-25 14:53:18 发布

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本文探讨了经典的三数之和为零问题，并提供了一种有效的解决方案。通过对输入数组进行排序并使用双指针技术，可以高效地找出所有唯一且不重复的三元组，这些三元组相加等于零。

Question:

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: 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]
]

Solution:

class Solution {
        public:
        vector<vector<int>> res;
        void find(vector<int>& v , int  s, int e , int data){
            while(s < e){
                int sum = v[s] + v[e] ; 
                if(sum == -data){
                    vector<int> tmp;
                    tmp.push_back(v[s]);
                    tmp.push_back(v[e]);
                    tmp.push_back(data);
                    res.push_back(tmp);
                    while(s != e && v[s] == v[s+1]){s++;}
                    while(e != 0 && v[e] == v[e-1]){e--;}
                    s++;
                    e--;
                }
                else if(sum < -data ){
                    s++;
                }
                else{
                    e--;
                }
            }
        }
        vector<vector<int>> threeSum(vector<int>& nums) {
            sort(nums.begin(),nums.end());
            int len = nums.size();
            for(int i = 0 ; i < len ; i++){
                if(i > 0 && nums[i] == nums[i-1])continue;
                find(nums,i+1,len-1,nums[i]);
            }
            return res;
        }  
    };