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这里能问个iterator超时的事吗 Time Limit Exceeded

2014/10/16镜像同步1 回复

是这样，leetcode上求4sum一道题， https://oj.leetcode.com/problems/4sum/ 我使用iterator的方式超时了，然后使用数组下标的方式居然不超时，不是说iterator更安全更好吗？或者有木有更好的方式呀？ [ema23] //这个超时了 class Solution { public: vector<vector<int> > fourSum(vector<int> &num, int target){ vector<vector<int> > res; int len = num.size(); if(len < 4) return res; sort(num.begin(), num.end()); for(vector<int>::const_iterator it = num.begin(); it != num.end(); it++){ if(it != num.begin() && *it == *(it-1)) continue; for(vector<int>::const_iterator it2 = it+1; it2 != num.end(); it2++){ if(it2 != it+1 && *it2 == *(it2-1)) continue; vector<int>::const_iterator front = it2+1; vector<int>::const_iterator back = num.end()-1; while(front < back){ if(front != it2+1 && *front == *(front-1)){ front++; continue; } if(back != num.end()-1 && *back == *(back+1)){ back--; continue; } int sum = *it + *it2 + *front + *back; if(sum < target) front++; else if(sum > target) back--; else{ vector<int> tmp; tmp.push_back(*it); tmp.push_back(*it2); tmp.push_back(*front); tmp.push_back(*back); res.push_back(tmp); front++; } } } } return res; } };

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