leetcode15---3Sum
2015-12-14 21:47
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问题描述:
如输入:
得到:
代码:
执行结果:
如果算例变为int a[]={-2,0,0,2,2};执行结果:
给定一个数组,在此数组中寻找出3个数,使得3个数的和为0;并且3个数按非递减顺序输出。
如输入:
{-1,0,1,2,-1,-4};
得到:
-1 -1 2 -1 0 1
代码:
#include <iostream> #include<vector> #include<algorithm> using namespace std; class Solution { public: vector<vector<int> > threeSum(vector<int>& nums) { int n=nums.size(); vector<vector<int> > result; sort(nums.begin(), nums.end());//快速排序 for(int i=0;i<n-2;i++) { if(i>0 && nums[i]==nums[i-1]) continue;//避免重复.在此之前已经寻找过。 /*###############此段代码快速寻找符合条件的两个数###################*/ for(int j=i+1,k=n-1;j<k;) { //if(nums[j]==nums[j+1]) j++;//避免重复.自动跳到下一个 //if(nums[k]==nums[k-1]) k--;//避免重复.自动跳到前一个 if(nums[j]+nums[k] == -nums[i]) { vector<int> tmp; tmp.push_back(nums[i]); tmp.push_back(nums[j]); tmp.push_back(nums[k]); result.push_back(tmp); tmp.clear();//每次清空tmp //if(nums[j]==nums[j+1]) j++;//避免重复.自动跳到下一个 //if(nums[k]==nums[k-1]) k--;//避免重复.自动跳到前一个 while(j<k && nums[j]==nums[j+1]) j++;//避免重复.自动跳到下一个 while(j<k && nums[k]==nums[k-1]) k--;//避免重复.自动跳到前一个 j++;//!!! k--;//!!! } else if(nums[j]+nums[k] < -nums[i]) j++; else k--; } /*###############此段代码快速寻找符合条件的两个数###################*/ } return result; } }; int main() { Solution s; int a[]={-4,-2,1,-5,-4,-4,4,-2,0,4,0,-2,3,1,-5,0};// int a2[]={-2,0,3,-1,4,0,3,4,1,1,1,-3,-5,4,0};// vector<int> vec(a,a+sizeof(a));//数组指针给vector赋值 //vector<int> vec(a2,a2+sizeof(a2));//数组指针给vector赋值 vector<vector<int> > res; //cout<<"begin"<<endl; res = s.threeSum(vec); //cout<<"end"<<endl; int n=res.size(); for(int i=0;i<n;i++) { for(int j=0;j<3;j++) { cout<<res[i][j]<<" "; } cout<<endl; } return 0; }
执行结果:
-5 1 4 -4 0 4 -4 1 3 -3 -1 4 -3 0 3 -2 -2 4 -2 -1 3 -2 1 1 -1 0 1 0 0 0 Process returned 0 (0x0) execution time : 0.054 s Press any key to continue.
如果算例变为int a[]={-2,0,0,2,2};执行结果:
-2 0 2 Process returned 0 (0x0) execution time : 0.109 s Press any key to continue.
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