[LeetCode 128] Longest Consecutive Sequence (Medium)

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

Clarification

Your algorithm should run in O(n) complexity.

Example

Given [100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.

思路

• 用全局变量count来记录每个连续sequence的长度
• 用全局变量maxLen来记录所有连续sequence的最长长度
• 先对数组排序
• 用2个指针，pre和cur

从1开始遍历数组

1. 需要去重复，比如[-1， -1， 0 ，0，1， 1， 2]

while(cur < num.length && num[cur] == num[cur - 1]) {
        pre++;
        cur++;
}

此处必须保证cur < num.length

2. 如果num[pre] + 1 = num[cur] 则说明是连续的sequence，更新pre, cur, maxLen
   此处必须保证cur < num.length
3. 如果不连续，那么重置count, 更新pre, cur

public class Solution {
    /*
     * @param num: A list of integers
     * @return: An integer
     */
    public int longestConsecutive(int[] num) {
        // write your code here
        if (num == null || num.length == 0) {
            return 0;
        }
        
        int pre = 0;
        int i = 1;
        
        int count = 1;
        int maxLen = 1;
        Arrays.sort(num);
        
        for (; i < num.length; i++) {
            while (i < num.length && num[i] == num[i - 1]) {
                i++;
                pre++;
            }
            if (i < num.length && num[pre] + 1 == num[i]) {
                pre++;
                count++;
                maxLen = count > maxLen ? count : maxLen;
            } else {
                count = 1;
                pre++;
            }
        }
        return maxLen;
    }
}