next-permutation.py

# Time:  O(n)
# Space: O(1)
#
# 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
#

class Solution:
    # @param {integer[]} nums
    # @return {void} Do not return anything, modify nums in-place instead.
    def nextPermutation(self, num):
        k, l = -1, 0
        for i in xrange(len(num) - 1):
            if num[i] < num[i + 1]:
                k = i
                
        if k == -1:
            num.reverse()
            return
        
        for i in xrange(k + 1, len(num)):
            if num[i] > num[k]:
                l = i
                
        num[k], num[l] = num[l], num[k]
        num[k + 1:] = num[:k:-1]

if __name__ == "__main__":
    num = [1, 4, 3, 2]
    Solution().nextPermutation(num)
    print num
    Solution().nextPermutation(num)
    print num
    Solution().nextPermutation(num)
    print num