15. 3Sum

原题目：

15. 3Sum

读题：

任意三个数相加为0，但是不能是相同的三元组

解题思路：

先排序，然后遍历序列，如果遇到相同的数则跳过，指定一个num[i]后，剩下的就是要找到另外两个数A+B=-num[i]，可以采用二分查找，begin = i+1;end = len -1;依次查找可以得到，以下是AC代码

class Solution {public:vector<vector<int>> threeSum(vector<int>& nums) {int i,j,k;int sum,begin,end;vector <vector<int>> result;int length = nums.size();if(length < 3){return result;}sort(nums.begin(),nums.end());for(i=0;i<length;i++){if(nums[i]>0){break;}if( i > 0 && nums[i] == nums[i-1]){continue;}begin = i+1;end = length-1;while(begin < end){sum = nums[i] + nums[begin] + nums[end];if (0 == sum){vector <int> temp;temp.push_back(nums[i]);temp.push_back(nums[begin]);temp.push_back(nums[end]);result.push_back(temp);begin++;end--;while(begin < end &&nums[begin] == nums[begin-1]) {begin++;}while(begin < end &&nums[end] == nums[end+1]) {end--;}}else if(sum>0){end--;}else {begin++;}}}return result;}};int main(){Solution s;int i,j;vector <vector <int>> result;vector <int> v;v.push_back(-1);v.push_back(0);v.push_back(1);v.push_back(2);v.push_back(-1);v.push_back(-4);result = s.threeSum(v);for(i=0;i<result.size();i++){for(j=0;j<result[i].size();j++){cout<<result[i][j]<<endl;}}}