leetcode--128--最长连续序列

题目：
给定一个未排序的整数数组 nums ，找出数字连续的最长序列（不要求序列元素在原数组中连续）的长度。

进阶：你可以设计并实现时间复杂度为 O(n) 的解决方案吗？

示例 1：

输入：nums = [100,4,200,1,3,2]
输出：4
解释：最长数字连续序列是 [1, 2, 3, 4]。它的长度为 4。
示例 2：

输入：nums = [0,3,7,2,5,8,4,6,0,1]
输出：9

提示：

0 <= nums.length <= 104
-109 <= nums[i] <= 109

链接：https://leetcode-cn.com/problems/longest-consecutive-sequence

Python代码：

class Solution(object):
    def longestConsecutive(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        size = len(nums)
        
        for i in range(size-1):
            for j in range(size-i-1):
                if nums[j]>nums[j+1]:
                    nums[j], nums[j+1] = nums[j+1], nums[j]

        ret = 1
        ans = 1
        for i in range(size-1):
            if (nums[i+1] == nums[i]+1):
                ans += 1
            elif (nums[i+1] == nums[i]):
                continue
            else:
                ans = 1
            ret = max(ret, ans)
        return ret

Python代码2：

class Solution(object):
    def longestConsecutive(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0

        ret = 0
        dt = {}

        for num in nums:
            if num in dt:
                continue
            left = dt.get(num-1, 0)
            right = dt.get(num+1, 0)

            curr = 1+left+right
            dt[num] = curr
            dt[num-left] = curr
            dt[num+right] = curr
            ret = max(ret, curr)
        return ret

Java代码：

class Solution {
    public int longestConsecutive(int[] nums) {
        Map<Integer, Integer> dt = new HashMap<>();
        if(nums.length==0){
            return 0;
        }
        int ret = 0;
        for(int num:nums){
            if(dt.containsKey(num)){
                continue;
            }
            int left = dt.getOrDefault(num-1, 0);
            int right = dt.getOrDefault(num+1, 0);
            int curr = 1 + left + right;
            dt.put(num, curr);
            dt.put(num-left, curr);
            dt.put(num+right, curr);
            ret = Math.max(ret, curr);
        }
        return ret;
    }
}