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leetcode : 3sum

Given an array S of n integers, are there elements abc 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)

 

思路还是先排序然后慢慢找,有意思的地方在于去重。
在得到一个和为0的i left 与 right 之后,再将附近的重复元素去掉就可以避免漏掉类似-2,1,1这样组合,并去掉重复组合
AC代码:
class Solution {public:vector<vector<int> > threeSum(vector<int> &num) {        vector<vector<int>> ret;        if(num.size() < 3)            return ret;        sort(num.begin(), num.end());        for(int i = 0; i < num.size() - 2; ++i){            int left = i + 1;            int right = num.size() - 1;            while(left < right){                if(num[left] + num[right] + num[i] > 0){                    --right;                }else if(num[left] + num[right] + num[i] < 0){                    ++left;                }else{                    ret.push_back(vector<int>{num[i], num[left], num[right]});                    while(right > left && num[right] == num[(right--) - 1])         //num[right]为最后一个重复元素,并且对right自减,left同理                        ;                    while(left < right && num[left] == num[(left++) +1])                        ;                }            }            while(i < num.size() - 1 && num[i + 1] == num[i])                ++i;        }        return ret;    }};

 

leetcode : 3sum