First Missing Positive

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

For example,
Given [1,2,0] return 3,
and [3,4,-1,1] return 2.

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

这题要求比较严格,要求O(n)时间,O(1)空间,所以简单的方法排序, hashmap等都无法使用.这里主流的思路是使用count sort, 达到 nums[0] = 1, nums[1] = 2,....这样的效果.

这样处理之后,再进行一次扫描, 第一个nums[i] != i+1的位置,就是first missing positive.代码如下:

class Solution(object):
    def firstMissingPositive(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        #count sort, nums[0] = 1, nums[1] =2, nums[2] = 3
        n = len(nums)
        for i in xrange(n):
            while nums[i] > 0 and nums[i] <= n and nums[nums[i] - 1] != nums[i]: #之所以不用nums[i] - 1 != i, 是为了防止两个位置的元素相等
                tmp = nums[nums[i] - 1]
                nums[nums[i] -1] = nums[i]
                nums[i] = tmp
                #nums[i],nums[nums[i]-1] = nums[nums[i]-1], nums[i] #注意这里直接交换是错误的
        for i in xrange(n):
            if nums[i] != i+1:
                return i+1
        return n+1

每次swap,都至少可以使一个元素回到正确位置,所以是O(n)时间复杂度.