【leetcode】1053. Previous Permutation With One Swap

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 <= A.length <= 10000

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 bisectdic = {}val_list = []for i in range(len(A)-1,-1,-1):inx = bisect.bisect_left(val_list,A[i])inx -= 1if inx >= 0 and inx < len(val_list):A[i], A[dic[val_list[inx]]] = A[dic[val_list[inx]]], A[i]breakif A[i] not in dic:bisect.insort_left(val_list,A[i])dic[A[i]] = ireturn A

转载于:https://www.cnblogs.com/seyjs/p/10929238.html

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