标签：des style blog color io ar strong for div
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.
For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
A solution set is:
(-1, 0, 0, 1)
(-2, -1, 1, 2)
(-2, 0, 0, 2)
思路：
1 class Solution { 2 public: 3 vector<vector<int>> fourSum( vector<int> &num, int target ) { 4 set<vector<int>> quadrupletSet; 5 sort( num.begin(), num.end() ); 6 vector<int> quadruplet( 4, -1 ); 7 for( int s1 = (int)num.size()-4; s1 >= 0; --s1 ) { 8 if( target < 4*num[s1] ) { continue; } 9 for( int e1 = (int)num.size()-1; e1 > s1+2; --e1 ) { 10 if( target > 4*num[e1] ) { break; } 11 int s2 = s1+1, e2 = e1-1; 12 while( s2 < e2 ) { 13 int sum = num[s1] + num[e1] + num[s2] + num[e2]; 14 if( sum == target ) { 15 quadruplet[0] = num[s1]; 16 quadruplet[3] = num[e1]; 17 quadruplet[1] = num[s2]; 18 quadruplet[2] = num[e2]; 19 quadrupletSet.insert( quadruplet ); 20 ++s2; --e2; 21 } else if( sum < target ) { 22 ++s2; 23 } else { 24 --e2; 25 } 26 } 27 } 28 } 29 vector<vector<int>> quadruplets; 30 for( auto iter = quadrupletSet.begin(); iter != quadrupletSet.end(); ++iter ) { 31 quadruplets.push_back( *iter ); 32 } 33 return quadruplets; 34 } 35 };
标签：des style blog color io ar strong for div
原文地址：http://www.cnblogs.com/moderate-fish/p/3959666.html
