Leetcode练习(Python)：数组类：第41题：给你一个未排序的整数数组，请你找出其中没有出现的最小的正整数。你的算法的时间复杂度应为O(n)，并且只能使用常数级别的额外空间。

Posted 2021-02-28 桌子哥

题目：给你一个未排序的整数数组，请你找出其中没有出现的最小的正整数。你的算法的时间复杂度应为O(n)，并且只能使用常数级别的额外空间。

思路：第一个思路是创建一个锚点，这个锚点表示第一个正整数的出现的位置，然后再分情况来判断，结果程序无法通过所有的测试用例，第一个思路方法以后再实现，后来使用HashMap来说实现，十分方便。

程序1：HashMap

class Solution:

    def firstMissingPositive(self, nums: List[int]) -> int:

        length = len(nums)

        hashmap = [0] * length

        for x in nums:

            if x > 0 and x <= length:

                hashmap[x - 1] = x      

        for index in range(length):

            if hashmap[index] != index + 1:

                return index + 1

        return length + 1

程序2：类似专家系统（此程序有问题，待修正）

class Solution:

    def firstMissingPositive(self, nums: List[int]) -> int:

        nums.sort()

        length = len(nums)

        if length <= 0:

            return 1

        if length == 1:

            if nums[0] <= 0:

                return 1

            elif nums[0] == 1:

                return 2

            else:

                return 1

        #Find the first positive integer as anchor

        for index in range(length):

            if nums[index] > 0:

                anchor = index

                break

            else:

                anchor = length - 1

        #temp_index is the first anchor

        #anchor in the head

        if anchor == 0:

            if nums[anchor] == 1:

                while anchor < length:

                    #anchor += 1

                    if nums[anchor] - nums[anchor - 1] > 1:

                        return nums[anchor - 1] + 1

                        break

                    elif nums[anchor] - nums[anchor - 1] == 1 and anchor - 1 < length:

                        anchor += 1

                    elif nums[anchor] - nums[anchor - 1] == 0 and anchor - 1 < length:

                        anchor += 1

                    if anchor >= length - 1:

                        anchor = length - 1

                        return nums[anchor] + 1

                        break

            else:

                return 1

            

        #anchor in the tail

        elif anchor == length - 1:

            if nums[anchor] < 1:

                return 1

            elif nums[anchor] > 1:

                return 1

            else:

                return nums[anchor] + 1

        #anchor in the body

        else:

            while anchor < length:

                anchor += 1

                if nums[anchor] - nums[anchor - 1] > 1:

                    return nums[anchor - 1] + 1

                    break

                elif nums[anchor] - nums[anchor - 1] == 1 and anchor + 2 < length:

                    anchor += 1

                elif nums[anchor] - nums[anchor - 1] == 0 and anchor + 2 < length:

                    anchor += 1

                if anchor >= length - 1:

                    anchor = length - 1

                    return nums[anchor] +  1

                    break

                #anchor += 1

        #return anchor