LeetCode----First Missing Positive

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.
分析：
寻找第一个Missing的正整数。

题意：题目需要给定的数组A在排序后满足：A[i] 等于 i+1 。

算法：不要使用基于比较的各种排序，可以使用计数排序和桶排序，我尝试了桶排序，虽然增加了额外的空间，但是题目要求并没有那么严格，通过了。

大空间消耗的代码：

class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None

class Solution(object):
def firstMissingPositive(self, nums):
"""
:type nums: List[int`]
:rtype: int
"""
SIZE = 40
DIVID = 10000
buckets = [None] * SIZE
for value in nums:
if value > 0:
index = value * SIZE / DIVID
self.insertToBucket(buckets, index, value)
is_first = True
res = []
for b in buckets:
if b:
while b:
res.append(b.val)
b = b.next
if len(res) == 0:
return 1
elif len(res) == 1:
if res[0] == 1:
return 2
else:
return 1
else:
pre = res[0]
if pre != 1:
return 1
for i in res[1:]:
if i != pre:
if i - pre != 1:
return pre + 1
else:
pre = i
if res[0] == 1:
return i + 1
else:
return res[0] - 1

def insertToBucket(self, buckets, index, value):
p = head = buckets[index]
if head is None:
buckets[index] = ListNode(value)
elif p and value < p.val:
buckets[index] = ListNode(value)
buckets[index].next = p
else:
while p and value >= p.val:
pre = p
p = p.next
if p is None:
pre.next = ListNode(value)
else:
pre.next = ListNode(value)
pre = pre.next
pre.next = p

def outputList(self, l):
l_p = l
while l_p:
print l_p.val,
l_p = l_p.next
print

O(1)空间的代码：

class Solution(object):
def firstMissingPositive(self, nums):
"""
:type nums: List[int`]
:rtype: int
"""
nums_len = len(nums)
if not nums_len:
return 1
for i in range(nums_len):
while nums[i] != i + 1:
if nums[i] <= 0 or nums[i] >= nums_len or nums[i] == nums[nums[i] - 1]:
break
k = nums[i]
nums[i], nums[k - 1] = nums[k - 1], nums[i]

for i, v in enumerate(nums):
if v != i + 1:
return i+1
return i + 2