寻找数组中的零和三元组-优快云博客

本文介绍了一种算法，用于在给定数组中找到三个元素，它们的和为零。通过将问题转化为两数之和问题并进行排序和去重处理，实现了高效查找。该方法适用于解决类似的数组中特定条件元素组合的问题。

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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)

solution : 转换为twosum问题，遍历数组，以当前数位target，求后续数组中和为-target的两数。

需要注意的地方是target需要跳过重复的数字，求twosum时也要跳过重复数字。

class Solution {
public:

    vector< vector<int> > threeSum(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int size = num.size();
        vector< vector<int> >result;
        vector<int>tri(3,0);
        int res = 0;
        if(size < 3)
            return result;
        
        sort(num.begin(),num.end());
        
        
        for(int index = 0; index < size-2; index++)
        {
            if(num[index] == num[index-1])continue;
            int res1 = -num[index];
            int start = index + 1;
            int end = size - 1;
            while(start < end)
            {
                int sum = num[start] + num[end];
                if(sum == res1)  //found 3 sum
                {
                    tri[0] = num[index];
                    tri[1] = num[start];
                    tri[2] = num[end];
                    result.push_back(tri);
                    while(num[start] == num[start+1])
                        start++;
                    start++;
                    end --;
                    res++;
                }
                else if(sum < res1)
                {
                    start ++;
                }
                else
                {
                    end --;
                }
            }
        }
        return result;
    }
};