【leetcode】15. 3Sum

题目

3Sum
Total Accepted: 92588 Total Submissions: 521219 Difficulty: Medium

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)

代码

#include <iostream>#include <vector>#include <algorithm>#include <map>#include <sys/time.h>using namespace std;vector<vector<int>> threeSum(vector<int>& nums) {  vector< vector<int> > ret;  if(nums.size()<3)    return ret;  std::sort( nums.begin(), nums.end() );  if(nums[0]>0||nums.back()<0)    return ret;  std::map<int, int> unmap;  for(int i=0;i<nums.size();++i) {    unmap.insert(map<int, int>::value_type(nums[i],i));  //重复键值,插入失败  }  for(int i=0;i<(nums.size()-2)&&nums[i]<=0;++i) {    for(int j=i+1;j<(nums.size()-1);++j) {      while(nums[j]==nums[j+1]) {        j++;      }      std::map<int,int>::const_iterator got = unmap.find( 0-nums[i]-nums[j] );      if( got!=unmap.end() &&( got->second > j) ) {        vector<int> a;        a.push_back(nums[i]);        a.push_back(nums[j]);        a.push_back(nums[got->second]);        ret.push_back( a );      }    }  }  std::sort(ret.begin(), ret.end());  auto iter = std::unique( ret.begin(), ret.end() );  ret.resize( std::distance( ret.begin(), iter ) );  return ret;}int main(int argc,char *argv[]) {  vector<int> nums={-9,14,-7,-8,9,1,-10,-8,13,12,6,9,3,-3,-15,-15,1,8,-7,-4,-6,8,2,-10,8,11,-15,3,0,-11,-1,-1,10,0,6,5,-14,3,12,-15,-7,-5,9,11,-1,1,3,-15,-5,11,-12,-4,-4,-2,-6,-10,-6,-6,0,2,-9,14,-14,-14,-9,-1,-2,-7,-12,-13,-15,-4,-3,1,14,3,-12,3,3,-10,-9,-1,-7,3,12,-6,0,13,4,-15,0,2,6,1,3,13,8,-13,13,11,11,13,14,-6};  struct  timeval    tv1,tv2;  struct  timezone   tz;  int time1 = gettimeofday(&tv1,&tz);  //func();  vector< vector<int> > ret = threeSum( nums );  int time2 = gettimeofday(&tv2,&tz);  cout << "time consuming:" << tv2.tv_usec - tv1.tv_usec <<"us"<< endl;  for(int i=0;i<ret.size();++i) {    for(int j=0;j<ret[0].size();++j) {      cout<<ret[i][j]<<",";    }    cout<<endl;  }  return 0;}

总结

1. 首先应对数组排序；
2. 注意跳过重复元素；
3. 对map的应用。