LeetCode 24 Longest Consecutive Sequence
2014-08-27 15:43
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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.
分析:
刚看还以为是最长连续子序列,第一反应是DP,于是尝试寻找最优子问题,发现并不是要求顺序的,那么应该没有那么复杂,
那么,处理没有顺序的一堆元素,首先想到hash哦,具体是HashSet还是HashMap就看要不要键值对咯。
这里,
1,把数组所有元素先放到HashSet里,
2,遍历数组元素,并尝试此元素左右连续的所有可能,直到碰见不连续。注意:碰见连续的,要从HashSet里删除。
是怎么保证O(n)复杂度的?
1,把数组所有元素放进HashSet是 n;
2,因为碰见set中访问过的,就会删除掉,所以保证set里的元素只访问了一遍,所以后面一个循环的复杂度是 2n, 即,遍历数组n,访问set 的元素 n。
所以整体复杂度是3n, 即 O(n).
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.
分析:
刚看还以为是最长连续子序列,第一反应是DP,于是尝试寻找最优子问题,发现并不是要求顺序的,那么应该没有那么复杂,
那么,处理没有顺序的一堆元素,首先想到hash哦,具体是HashSet还是HashMap就看要不要键值对咯。
这里,
1,把数组所有元素先放到HashSet里,
2,遍历数组元素,并尝试此元素左右连续的所有可能,直到碰见不连续。注意:碰见连续的,要从HashSet里删除。
是怎么保证O(n)复杂度的?
1,把数组所有元素放进HashSet是 n;
2,因为碰见set中访问过的,就会删除掉,所以保证set里的元素只访问了一遍,所以后面一个循环的复杂度是 2n, 即,遍历数组n,访问set 的元素 n。
所以整体复杂度是3n, 即 O(n).
public class Solution { public int longestConsecutive(int[] num) { Set<Integer> set = new HashSet<Integer>(); int max = 1; for(int e : num) set.add(e); for(int e : num){ int left = e-1; int right = e+1; int count = 1; while(set.contains(left)){ count++; set.remove(left); left--; } while(set.contains(right)){ count++; set.remove(right); right++; } max = Math.max(count, max); } return max; } }
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