[Leetcode][Python]31: Next Permutation

最新推荐文章于 2021-03-10 11:29:57 发布

转载最新推荐文章于 2021-03-10 11:29:57 发布 · 171 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Dabay/p/4263551.html

文章标签：

#python

本文详细解析了LeetCode第31题“下一排列”的算法实现，该问题要求在给定数字序列的基础上找到下一个字典序更大的排列。通过具体示例，如将598421转换为812459的过程，阐述了解决方案的步骤：首先从末位开始寻找升序结束的位置，然后交换关键位置的数字并反转后续子序列。

# -*- coding: utf8 -*-
'''
__author__ = 'dabay.wang@gmail.com'

31: Next Permutation
https://oj.leetcode.com/problems/next-permutation/

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

===Comments by Dabay===

比如 5 9 8 4 2 1
从末尾1开始往前面看，知道升序结束。在【9 8 4 2 1】中找一个正好比5大的数8，交换变成8 9 5 4 2 1。
然后把 【9 5 4 2 1】反转，变成【1 2 4 5 9】。最后结果为： 8 1 2 4 5 9


'''

class Solution:
    # @param num, a list of integer
    # @return a list of integer
    def nextPermutation(self, num):
        i = len(num) - 1
        while i > 0:
            if num[i-1] >= num[i]:
                i -= 1
                continue
            j = i
            while j < len(num):
                if num[j] > num[i-1]:
                    j += 1
                    continue
                break
            num[i-1], num[j-1] = num[j-1], num[i-1]
            tails = num[i:]
            tails.reverse()
            return num[:i] + tails
        else:
            num.reverse()
            return num


def main():
    s = Solution()
    nums = [5, 1, 1]
    print s.nextPermutation(nums)


if __name__ == "__main__":
    import time
    start = time.clock()
    main()
    print "%s sec" % (time.clock() - start)