leetcod:最长连续序列(python)

乖乖的函数 2020-11-13 02:57:14
Python 最长 序列 连续 leetcod


1. 题目描述

给定一个未排序的整数数组,找出最长连续序列的长度。
要求算法的时间复杂度为 O(n)。

示例:

输入: [100, 4, 200, 1, 3, 2]
输出: 4
解释: 最长连续序列是 [1, 2, 3, 4]。它的长度为 4。

2. 思路

参考

https://leetcode-cn.com/problems/longest-consecutive-sequence/solution/zui-chang-lian-xu-xu-lie-by-leetcode/

2.1 暴力穷举

遍历nums 数组中的每一个数字,并将该数字作为连续序列的第一个数字,枚举后面的数字,直到有数字在原数组中没出现过。当枚举到数组中没有的数字时,记录下序列的长度,并更新当前最优解。

会超时

class Solution:
def longestConsecutive(self, nums: List[int]) -> int:
res = 0
for i in nums:
count = 0
j = i
while j in nums:
count += 1
j += 1
res = max(res,count)
return res

2.2 先排序,再找

时间复杂度为o(nlogn)
class Solution:
def longestConsecutive(self, nums: List[int]) -> int:
if nums == []:
return 0
nums.sort()
maxlen = 1
curlen = 1
for i in range(1,len(nums)):
if nums[i] != nums[i-1]: # 如果有重复的数字,长度算1
if nums[i] == nums[i-1]+1:
curlen += 1
maxlen = max(maxlen,curlen)
else:
maxlen = max(maxlen,curlen)
curlen = 1
return maxlen

2.3 利用set

空间换时间,借助一个set集合,对暴力方法进行优化。

class Solution:
def longestConsecutive(self, nums: List[int]) -> int:
if len(nums) == 0:
return 0
numset = set(nums)
maxlen = 0
for num in numset:
curlen = 1
if num - 1 not in numset:
# 只有当num是一个序列的端点时,在进行循环,复杂度为o(n+n)= o(n)
while num + 1 in numset:
curlen += 1
num += 1
maxlen = max(maxlen,curlen)
return maxlen
版权声明
本文为[乖乖的函数]所创,转载请带上原文链接,感谢
https://blog.csdn.net/ggdhs/article/details/91364477

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