leetcode:Longest Consecutive Sequence
2014-09-17 09:48
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题目:https://oj.leetcode.com/problems/longest-consecutive-sequence/
Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
For example,
Given
The longest consecutive elements sequence is
length:
Your algorithm should run in O(n) complexity.
题目意思:找出最长连续序列;
要求:在O(N)时间复杂度下完成。
一开始的想法当然是先对数组进行排序,然后遍历一遍完成。但是最好的排序算法的复杂度也是O(nlogn),显然不符合题目要求就没有进行尝试,但是在网上看到有人说这样做也是可以提交通过的。
后来的想法是,对于数组中每一个元素n,递归查找n-1是否在数组中,n+1是否在数组中...计算最长值。但是要查找一个数是否在数组中,先要对数组进行一下预处理。想要实现O(1)时间的查找速度,必须得想到哈希表了~在C++中,就是unordered_set.
AC代码:
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.
题目意思:找出最长连续序列;
要求:在O(N)时间复杂度下完成。
一开始的想法当然是先对数组进行排序,然后遍历一遍完成。但是最好的排序算法的复杂度也是O(nlogn),显然不符合题目要求就没有进行尝试,但是在网上看到有人说这样做也是可以提交通过的。
后来的想法是,对于数组中每一个元素n,递归查找n-1是否在数组中,n+1是否在数组中...计算最长值。但是要查找一个数是否在数组中,先要对数组进行一下预处理。想要实现O(1)时间的查找速度,必须得想到哈希表了~在C++中,就是unordered_set.
AC代码:
class Solution { public: int longestConsecutive(vector<int> &num) { //为了保证O(1)时间的查找速度,先建立哈希表存储数组元素 unordered_set<int> numSet; for(int i = 0;i < num.size();i++){ numSet.insert(num[i]); } int max = 1; //遇到一个数n,递归查找n-1和n+1是否在数组中... for(int i = 0;i < num.size();i++){ int left = num[i] - 1; int right = num[i] + 1; int cur = 1; //一个一个向左查找 while(numSet.count(left)){ cur++; //每找到一个就从哈希表中删除掉,避免重复查找,一个数只会出现在一个连续序列中 numSet.erase(left); left--; } while(numSet.count(right)){ cur++; numSet.erase(right); right++; } max = cur > max ? cur : max; } return max; } };
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