标签：style blog io strong for ar div log amp
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.
两种方法：1. 利用 hash_map 结构，数组有序时查找的思想。
class Solution {
public:
int longestConsecutive(vector<int> &num) {
if(num.size() == 0) return 0;
unordered_map<int, bool> _map;
for(size_t i = 0; i < num.size(); ++i)
_map.insert(pair<int ,bool>(num[i], false));
int ans = 0;
for(auto it = _map.begin(); it != _map.end(); ++it) {
if(it->second) continue; // visited
int len = 1;
int v1 = it->first-1;
while(_map.count(v1)) { _map[v1--] = true; len++; }
int v2 = it->first+1;
while(_map.count(v2)) { _map[v2++] = true; len++; }
if(len > ans) ans = len;
}
return ans;
}
};
2. 动态的构造有向线段（矢量）(两端为线段始末位置)。若当前点可增加有向线段长度，拓展线段。
class Solution {
public:
int longestConsecutive(vector<int> &num) {
int answer = 0;
unordered_map<int ,int>_map;
int low, high;
for(int i = 0; i < num.size(); ++i) {
if(_map.count(num[i])) continue; // must be used.
_map[num[i]] = num[i];
low = high = num[i];
if(_map.count(num[i]-1)) low = _map[num[i]-1];
if(_map.count(num[i]+1)) high = _map[num[i]+1];
answer = max(answer, high - low + 1);
if(low == high) continue;// not in a line
_map[low] = high;
_map[high] = low;
}
return answer;
}
};
24. Longest Consecutive Sequence
标签：style blog io strong for ar div log amp
原文地址：http://www.cnblogs.com/liyangguang1988/p/3938543.html
