题目如下：

Given an array A of positive integers (not necessarily distinct), return the lexicographically largest permutation that is smaller than A, that can be made with one swap (A swap exchanges the positions of two numbers A[i] and A[j]).  If it cannot be done, then return the same array.

 

Example 1:

Input: [3,2,1]Output: [3,1,2]Explanation: Swapping 2 and 1.

Example 2:

Input: [1,1,5]Output: [1,1,5]Explanation: This is already the smallest permutation.

Example 3:

Input: [1,9,4,6,7]Output: [1,7,4,6,9]Explanation: Swapping 9 and 7.

Example 4:

Input: [3,1,1,3]Output: [1,3,1,3]Explanation: Swapping 1 and 3.

 

Note:

1. 1 <= A.length <= 10000
2. 1 <= A[i] <= 10000

解题思路：要找出字典序小于自己的最大值，方法如下：从后往前遍历A，对于任意一个A[i]，在[i+1,A.length]区间内找出比自己小的最大值，如果能找到这样的值，则这两个元素交换，交换之后的A即为字典序小于自己的最大值。怎么找出[i+1,A.length]区间内找出比自己小的最大值？可以把区间内所有的值存入有序的数组中，通过二分查找即可。

代码如下：

class Solution(object):    def prevPermOpt1(self, A):        """        :type A: List[int]        :rtype: List[int]        """        import bisect        dic = {}        val_list = []        for i in range(len(A)-1,-1,-1):            inx = bisect.bisect_left(val_list,A[i])            inx -= 1            if inx >= 0 and inx < len(val_list):                A[i], A[dic[val_list[inx]]] = A[dic[val_list[inx]]], A[i]                break            if A[i] not in dic:                bisect.insort_left(val_list,A[i])            dic[A[i]] = i        return A