标签：des style blog color io ar strong for div
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)
思路：
1 class Solution { 2 public: 3 vector<vector<int> > threeSum( vector<int> &num ) { 4 vector<vector<int> > triplets; 5 sort( num.begin(), num.end() ); 6 int size = num.size(); 7 vector<int> triplet( 3, -1 ); 8 for( int i = 0; i < size-2; ++i ) { 9 if( i > 0 && num[i] == num[i-1] ) { continue; } 10 if( num[i] > 0 ) { break; } 11 int s = i+1, e = size-1; 12 while( s < e ) { 13 if( num[e] < 0 ) { break; } 14 if( s > i+1 && num[s] == num[s-1] ) { ++s; continue; } 15 if( e < size-1 && num[e] == num[e+1] ) { --e; continue; } 16 int sum = num[i] + num[s] + num[e]; 17 if( sum == 0 ) { 18 triplet[0] = num[i]; 19 triplet[1] = num[s]; 20 triplet[2] = num[e]; 21 triplets.push_back( triplet ); 22 ++s; --e; 23 } else if( sum < 0 ) { 24 ++s; 25 } else { 26 --e; 27 } 28 } 29 30 } 31 return triplets; 32 } 33 };
标签：des style blog color io ar strong for div
原文地址：http://www.cnblogs.com/moderate-fish/p/3960253.html
