Leetcode - Longest Consecutive Sequence-优快云博客

本文介绍了一种寻找数组中最长连续序列的有效算法。提供了两种方法：一种是对数组进行排序后扫描，另一种是使用哈希集合实现O(N)的时间复杂度。讨论了如何避免重复元素的影响，并给出了详细的代码实现。

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[分析]
base version说几句：
[color=red]数组题一定要考虑重复重复重复的问题[/color]！另外循环结束要记得最后一次更新maxLen
O(N)解法思路请移步[url]http://blog.youkuaiyun.com/linhuanmars/article/details/22964467[/url]


public class Solution {
    // Method 1: base version, first sort, then scan to find
    public int longestConsecutive1(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        Arrays.sort(nums);
        int len = 1, maxLen = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] == nums[i - 1])
                continue;
            if (nums[i] == nums[i - 1] + 1) {
                len++;
            } else {
                maxLen = Math.max(maxLen, len);
                len = 1;
            }
        }
        maxLen = Math.max(maxLen, len);
        return maxLen;
    }
    // Method 2: O(n), use hashset
    public int longestConsecutive(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        HashSet<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < nums.length; i++) {
            set.add(nums[i]);
        }
        int maxLen = 0;

        while (!set.isEmpty()) {
            Iterator<Integer> iter = set.iterator();
            int item = iter.next();
            set.remove(item);
            int len = 1;
            int i = item - 1;
            while (set.contains(i)) {
                set.remove(i--);
                len++;
            }
            i = item + 1;
            while (set.contains(i)) {
                set.remove(i++);
                len++;
            }
            if (len > maxLen)
                maxLen = len;
        }
        return maxLen;
    }
}