250分的题，浪费太多时间了，这题就没太多时间看了。这道题初看起来没什么思路，但仔细看看给出的字串还是有规律可循的
Problem Statement
The signature of a permutation is a string that is computed as follows: for each pair of consecutive elements of the permutation, write down the letter 'I' (increasing) if the second element is greater than the first one, otherwise write down the letter 'D' (decreasing).
For example, the signature of the permutation {3,1,2,7,4,6,5} is "DIIDID".
Your task is to reverse this computation: You are given a string signature containing the signature of a permutation. Find and return the lexicographically smallest permutation with the given signature. If no such permutation exists, return an empty int[] instead.
Definition
Class: PermutationSignature
Method: reconstruct
Parameters: string
Returns: int[]
Method signature: int[] reconstruct(string signature)
(be sure your method is public)
Notes
- For any positive integer N, a permutation of N elements is a sequence of length N that contains each of the integers 1 through N exactly once.
- To compare two permutations A and B, find the smallest index i such that A[i] and B[i] differ. If A[i] < B[i], we say that A is lexicographically smaller than B, and vice versa.
Constraints
- signature will contain between 1 and 50 characters, inclusive.
- Each character in signature will be either 'I' or 'D'.
Examples
0)
"IIIII"
Returns: {1, 2, 3, 4, 5, 6 }
1)
"DI"
Returns: {2, 1, 3 }
There are two permutations with this signature: {3,1,2} and {2,1,3}. You must return the lexicographically smaller one.
2)
"IIIID"
Returns: {1, 2, 3, 4, 6, 5 }
3)
"DIIDID"
Returns: {2, 1, 3, 5, 4, 7, 6 }
分析
给一个string signature，signature的每个元素为‘I’或‘D’，让你构造出字典序最小的排列p[]，满足sign[i]为‘I’表示p[i] < p[i + 1]，sign[i]为'D'表示p[i] > p[i + 1]。
算法1
依次处理每个a[0 ~ n]，有如下两种情况：
1）当signature[i] == 'I'时，有a[i + 1] = i + 2
2）当signature[i] == 'D'时，就把a最后一段递减数列全部增加1后再让a[i + 1] = a[i] - 1
signature= "IDDIID"
p的变化过程是这样的：
1
12
132
1432（把32都增加1变为43，再添加一个2）
14325
143256
1432576（把6增加1变为7，再添加一个6）
这样每一步都能保证是当前字典序最小的排列，所以最终答案也是字典序最小的了。
public int[] reconstruct(string signature) { int len = signature.Length; int[] a = new int[len + 1]; a[0] = 1; for (int i = 0; i < len; i++) { if (signature[i] == 'D') { int j = i; int last = a[j]; a[j]++; while (--j >= 0 && a[j] > last) { last = a[j]; a[j]++; } a[i + 1] = a[i] - 1; } else { a[i + 1] = i + 2; } } return a; }
参考：http://www.cnblogs.com/zbwmqlw/archive/2011/02/11/1951251.html
算法2
很巧妙的算法，从1开始递增顺序依次填数
public int[] reconstruct(string signature) { int len = signature.Length; int[] a = new int[len + 1]; int i, j; for (i = 1; i <= len + 1; i++) { for (j = 0; j < len; j++) { if (a[j] == 0 && (signature[j] == 'I' || a[j + 1] != 0)) break; } a[j] = i; } return a; }
参考：chokudai的代码
