class Solution { public: int longestConsecutive(vector<int> &num) { int len = num.size(); int max_cons = 0; int cur_cons = 0; unordered_map<int, int> sgn; unordered_map<int, int>::iterator iter; for (int i=0; i<len; i++) { sgn.insert(make_pair(num[i], 0x1)); } for (int i=0; i<len; i++) { iter = sgn.find(num[i]); if (iter == sgn.end()) continue; // illegal case, should not hit if (iter->second == 0) continue; // this range has been scaned iter->second = 0; cur_cons = 1; int try_n = num[i]; // search towards negative while (try_n != INT_MIN) { try_n--; iter = sgn.find(try_n); if (iter == sgn.end()) break; iter->second = 0; cur_cons++; } try_n = num[i]; // search towards positive while (try_n != INT_MAX) { try_n++; iter = sgn.find(try_n); if (iter == sgn.end()) break; iter->second = 0; cur_cons++; } if (cur_cons > max_cons) max_cons = cur_cons; } return max_cons; } };
最直观的方法肯定是排序一下,然后从前到后扫描一遍即可,不过排序要nlogn时间,要在O(n)时间内完成的话,肯定不是基于比较排序了。可以使用桶排序,但是int的范围还是很大的,不可取。最后采用稀疏的表示方式,就是放入hash表中,第一次遍历数组插入以元素为key,1为value的hash表项,第二次遍历时尝试对每个元素的前驱后继值在hash表中进行一次查找,如果存在就继续向前或向后查找,更新连续计数cur_cons,同时将扫描到的连续hash表项的value值置零以表示该项已经进行了扫描,这样可以避免重复检测。
LeetCode Longest Consecutive Sequence,布布扣,bubuko.com
LeetCode Longest Consecutive Sequence
原文地址:http://www.cnblogs.com/lailailai/p/3813717.html