【ATT】Longest Consecutive Sequence
Q:
Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
For example,
Given [100, 4, 200, 1, 3, 2]
,
The longest consecutive elements sequence is [1, 2, 3, 4]
. Return its length: 4
.
Your algorithm should run in O(n) complexity.
A:使用Map来处理。对cluster来说,最重要的是上界,下界和长度。
对于a[i],只需要检查a[i]-1,a[i]+1
映射上界下界到区间长度。merge 相邻的cluster时,只需要更新上界和下界对应的区间长度。
//1. The key factors about a cluster is: lowest, highest, and length. //2. Map lowest and highest to length. To merge two neighbor clusters, only need to update it's new lowest and highest, with new length. //3. For every a[i], checking its neighbor a[i]-1 and a[i]+1 is enough. int longestConsecutive(vector<int> &num) { // Start typing your C/C++ solution below // DO NOT write int main() function if(num.empty()) return 0; int maxlen = 1; //注意,初始化时1,而不是0 map<int,int> hmap; for(int i=0;i<num.size();i++) { if(hmap.find(num[i])!=hmap.end()) //duplidate continue; hmap[num[i]] = 1; if(hmap.find(num[i]-1)!=hmap.end()) maxlen = max(maxlen,merge(hmap,num[i]-1,num[i])); if(hmap.find(num[i]+1)!=hmap.end()) maxlen = max(maxlen,merge(hmap,num[i],num[i]+1)); } return maxlen; } int merge(map<int,int>& hmap,int left,int right) { int lower = left-hmap[left]+1;//新区间的下届 int upper = right+hmap[right]-1; //新区间的上届 int len = upper - lower +1; hmap[lower] = len; hmap[upper] = len; return len; }