## 题目描述：

LeetCode 484. Find Permutation

By now, you are given a secret signature consisting of character 'D' and 'I'. 'D' represents a decreasing relationship between two numbers, 'I' represents an increasing relationship between two numbers. And our secret signature was constructed by a special integer array, which contains uniquely all the different number from 1 to n (n is the length of the secret signature plus 1). For example, the secret signature "DI" can be constructed by array [2,1,3] or [3,1,2], but won't be constructed by array [3,2,4] or [2,1,3,4], which are both illegal constructing special string that can't represent the "DI" secret signature.

On the other hand, now your job is to find the lexicographically smallest permutation of [1, 2, ... n] could refer to the given secret signature in the input.

Example 1:

```Input: "I"
Output: [1,2]
Explanation: [1,2] is the only legal initial spectial string can construct secret signature "I", where the number 1 and 2 construct an increasing relationship.
```

Example 2:

```Input: "DI"
Output: [2,1,3]
Explanation: Both [2,1,3] and [3,1,2] can construct the secret signature "DI",
but since we want to find the one with the smallest lexicographical permutation, you need to output [2,1,3]
```

Note:

• The input string will only contain the character 'D' and 'I'.
• The length of input string is a positive integer and will not exceed 10,000

## 题目大意：

• 输入字符串只包含字符'D'和'I'。
• 输入字符串的长度不会超过10000。

## 解题思路：

```初始令数组nums = [1,2, ..., n]，令数组ans = []

记s中的当前字符为c

若c == 'I'，则直接将nums中的最小元素移除并加入ans；将c从s中移除

否则，记连续的字符'D'的个数为cnt，将nums[0 ... cnt+1]移除，逆置后加入ans；将cnt个'D'从s中移除，如果后面有字符'I'，则一并移除。```

## Python代码：

``````class Solution(object):
def findPermutation(self, s):
"""
:type s: str
:rtype: List[int]
"""
size = len(s)
nums = list(range(1, size + 2))
ans = []
idx = 0
while idx < size:
if s[idx] == 'D':
cnt = 0
while idx < size and s[idx] != 'I':
idx += 1
cnt += 1
ans += nums[:cnt+1][::-1]
nums = nums[cnt+1:]
if idx < size:
idx += 1
else:
ans += [nums]
nums = nums[1:]
idx += 1
return ans + nums
`````` Pingbacks已关闭。

1. 赤壁的火神 发布于 2017年1月24日 08:39 #

这道题我想问一个Python的问题。。。第8行的代码改成我这样的：nums = [i for i in range(1, size + 2)]为什么最后会是一个顺序数组。。。我自己在VS上跑结果都是对的，怎么在OJ上就一直错误。。。难以理解。。。