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.
题目要求我们用O(n)的时间复杂度来解决,我们可以采用并查集的思想,当扫描到一个元素,我们从当前元素的两边开始处理,如果连续的都进行标记,同时维护一个最大值,再次扫描的时候如果被标记过的就可以跳过,这样我们就可以在O(n)的时间复杂度下找到最长的连续序列。代码如下:
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.
题目要求我们用O(n)的时间复杂度来解决,我们可以采用并查集的思想,当扫描到一个元素,我们从当前元素的两边开始处理,如果连续的都进行标记,同时维护一个最大值,再次扫描的时候如果被标记过的就可以跳过,这样我们就可以在O(n)的时间复杂度下找到最长的连续序列。代码如下:
public class Solution {
public int longestConsecutive(int[] nums) {
if(nums == null) return 0;
HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
int max = 1;
for(int i : nums) hm.put(i, 0);
for(int i = 0; i < nums.length; i++) {
int count = 1;
int left = -1;
int right = 1;
if(hm.get(nums[i]) == 1) continue;
hm.put(nums[i], 1);
while(hm.containsKey(nums[i] + left)) {
hm.put(nums[i] + left, 1);
count ++;
left --;
}
while(hm.containsKey(nums[i] + right)) {
hm.put(nums[i] + right, 1);
count ++;
right ++;
}
max = Math.max(max, count);
}
return max;
}
}