原题链接在这里：https://leetcode.com/problems/find-permutation/description/
题目：
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
题解：
先生成sorted array. 若出现连续的D时就把连续D开始和结尾对应的这段reverse.
Time Complexity: O(n). n = s.length().
Space: O(1). regardless res.
AC Java:
1 class Solution {
2 public int[] findPermutation(String s) {
3 int len = s.length();
4 int [] res = new int[len+1];
5 for(int i = 0; i<len+1; i++){
6 res[i] = i+1;
7 }
8
9 for(int i = 0; i<len; i++){
10 if(s.charAt(i) == 'D'){
11 int mark = i;
12 while(i<len && s.charAt(i)=='D'){
13 i++;
14 }
15 reverse(res, mark, i);
16 }
17 }
18
19 return res;
20 }
21
22 private void reverse(int [] res, int i, int j){
23 while(i<j){
24 swap(res, i++, j--);
25 }
26 }
27
28 private void swap(int [] res, int i, int j){
29 int temp = res[i];
30 res[i] = res[j];
31 res[j] = temp;
32 }
33 }
