題目：
Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
The replacement must be in-place, do not allocate extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1
分析：如果數(shù)組從大到小有序胀茵，則當(dāng)前為最大擂啥，只需轉(zhuǎn)為從小到大排序即可。如果不是腌零，那么此時需從右至左掃描找到第一個num[i]>nums[i-1]的下標(biāo)眯娱，在i-1右邊選出比nums[i-1]大的最小的數(shù)交換航揉，然后從小到大將i-1之后的數(shù)字排序即可浦夷。
代碼：
def pop_sort(self,nums,l,r):
for i in range(0,r-l):
for j in range(l,r-i):
if nums[j] > nums[j+1]:
nums[j],nums[j+1] = nums[j+1],nums[j]
def nextPermutation(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
n = len(nums)
i = n-1
while i>0 and nums[i]<=nums[i-1]:
i -= 1
if i == 0:
self.pop_sort(nums,0,n-1)
else:
j = n-1
while nums[j] <= nums[i-1]:
j -= 1
nums[j],nums[i-1] = nums[i-1],nums[j]
self.pop_sort(nums,i,n-1)
