[leetcode]128. Longest Consecutive Sequence最长连续序列
Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
Your algorithm should run in O(n) complexity.
Example:
Input: [100, 4, 200, 1, 3, 2]
Output: 4
Explanation: The longest consecutive elements sequence is[1, 2, 3, 4]
. Therefore its length is 4.
题目
给定一个数组,计算将其排序以后能形成的最长连续序列。
思路
如果允许O(nlogn)的复杂度,那么可以先排序。 以下是naive 的先排序的代码实现
class Solution {
public int longestConsecutive(int[] nums){
Arrays.sort(nums);
if(nums == null || nums.length == 0) return 0;
int result = 1;
int length = 1; for (int i = 1; i < nums.length; i++ ) {
if(nums[i] == nums[i-1] + 1 ){
length ++;
}else if (nums[i] == nums[i-1]){
continue;
}else{
length = 1; }
result = Math.max(result, length);
} return result;
}
}
再思考:
可是本题要求O(n)。
由于序列里的元素是无序的,又要求O(n),想到用哈希set。
代码
public int longestConsecutive(int[] nums) {
if(nums.length == 0) return 0;
Set<Integer> set = new HashSet<Integer>();
int max = 1;
for(int num : nums)
set.add(num);
for(int i = 0; i < nums.length; i++){
if(!(set.contains(nums[i] - 1))){
int currentSequence = 0;
int next = nums[i];
while(set.contains(next)){
currentSequence++;
max = Math.max(max, currentSequence);
next++;
}
}
} return max;
}
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