一.問題描述
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.
給定一組integers,找出其中最小的缺失的正整數值,要求時間複雜度爲O(n),空間複雜度爲O(1)。
二.代碼編寫
時間複雜度爲O(n)意味着不能直接對list進行排序O(NlogN),空間複雜度爲常數意味着不能新建一個list。
常數空間,我們應該想到直接swap,將相應得到數字m放到list的(m-1)的位置上。全部交換完畢後,返回list中第一個不滿足nums[m]!=m+1的(m+1)值。
代碼如下:
'''
@2016.11.20 by wttttt
@ for problem details, see https://leetcode.com/problems/first-missing-positive/
@ for solution description,see
@ trick
'''
class Solution(object):
def firstMissingPositive(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if nums == []:
return 1
len_nums = len(nums)
i = 0
while i< len_nums:
if nums[i]!=i+1 and nums[i]<=len_nums and nums[i]>0:
#nums[i],nums[nums[i]-1] = nums[nums[i]-1],nums[i] # swap
tmp = nums[i]
if tmp == nums[tmp-1]:
i += 1
continue
nums[i] = nums[tmp-1]
nums[tmp-1]=tmp
else:
i += 1
for i in range(len_nums):
if nums[i] != i+1:
return i+1
return nums[-1]+1