LeetCode 41. First Missing Positive--Python 解法--数学题-找到不存在的最小正整数-O(1)空间复杂度

题目地址：First Missing Positive - LeetCode

Given an unsorted integer array, find the smallest missing positive integer.

Example 1:

Input: [1,2,0]
Output: 3

Example 2:

Input: [3,4,-1,1]
Output: 2

Example 3:

Input: [7,8,9,11,12]
Output: 1

Note:

Your algorithm should run in O(n) time and uses constant extra space.

这道题目跟这道很像：LeetCode 31. Next Permutation-- Python 解法–数学题–比当前数大的最小的数

但这道题目难度明显更大，因为没有明确数组大小，而且有负数。

首先，时间复杂度至少为O(N)，因为要遍历数组一遍，空间复杂度可大可小，但O(N)的复杂度是可以接受的。

首先确定数组长度，最小缺失的正整数最大为N+1

遍历数组，使用bitmap确认存这个数是否存在过

Python解法如下：

class Solution:def firstMissingPositive(self, nums) -> int:length = len(nums)l = [0]*(length+2)for i in nums:if 0 < i < 1 + length:l[i] = 1for i in range(1, length+3):if l[i] == 0:return i

此解法时间复杂度为O(n)，空间复杂度为O(n)。

想要空间复杂度降为O(1)是可以做到的，但不一定会更快，因为CPU缓存的原因，到处交换数组不一定快：

class Solution:def firstMissingPositive(self, nums) -> int:end = len(nums)i = 0while i != end:val = nums[i]if val == i+1:i += 1elif val > end or val <= 0 or nums[val-1] == val:end -= 1nums[i], nums[end] = nums[end], nums[i]else:nums[i], nums[val-1] = nums[val-1], nums[i]return i+1

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