标签：des style blog http color io os ar 使用
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:
思路：先对数组排序，之后三个变量i(0->num.size()-1),j(initial:i+1),k(initial:num.size()-1)，i 扫描数组，判定sum=(num[i]+num[j]+num[k]==0) ? OK:(sum>0 ? --k:++j)；
问题的关键在于细节的考虑，若去处一下程序中的三个if判断，则对于上述example，（-1，-1,2）将有两个输出，判断在于对重复情况的去除
code:
class Solution { public: vector<vector<int> > threeSum(vector<int> &num) { vector<vector<int> >result; if(num.size()<3) return result; sort(num.begin(),num.end()); int sum=0; vector<int>tmp; for(int i=0;i<num.size();++i) { if(i>0&&num[i]==num[i-1]) continue; int j=i+1; int k=num.size()-1; while(j<k) { if(j>i+1&&num[j]==num[j-1]) { ++j; continue; } if(k<num.size()-1&&num[k]==num[k+1]) { --k; continue; } sum=num[i]+num[j]+num[k]; if(sum<0) ++j; else if(sum>0) --k; else { tmp.push_back(num[i]); tmp.push_back(num[j]); tmp.push_back(num[k]); result.push_back(tmp); tmp.clear(); ++j; --k; } } } return result; } };
也可以使用STL中相关函数解决上面问题
prev:
template <class BidirectionalIterator>
BidirectionalIterator prev (BidirectionalIterator it,
typename iterator_traits<BidirectionalIterator>::difference_type n = 1);
-n positions.If it is a random-access iterator, the function uses just once operator+ or operator-. Otherwise, the function uses repeatedly the increase or decrease operator (operator++ or operator--) on the copied iterator until n elements have been advanced.
upper_bound(first,last,x):返回值是寻找到元素下一位置的迭代器；
code:
class Solution { public: vector<vector<int> > threeSum(vector<int> &num) { vector<vector<int> >result; if(num.size()<3) return result; sort(num.begin(),num.end()); auto last=num.end(); for(auto a=num.begin();a<prev(last);a=upper_bound(a,prev(last),*a)) { for(auto b=next(a);b<prev(last);b=upper_bound(b,prev(last),*b)) { const int c=-(*a+*b); if(binary_search(next(b),last,c)) result.push_back(vector<int> {*a,*b,c}); } } return result; } };
标签：des style blog http color io os ar 使用
原文地址：http://www.cnblogs.com/chengyuz/p/4050831.html
