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

假设字符串 IDIIDD 初始化为 1234567
对于连续的 k 个连续的 D s，只需要翻转 [i, i+k] 的数.

IDIIDD1234567 // sorted1324765 // answer

代码如下：

public class Solution {    public int[] findPermutation(String s) {        char[] chs = s.toCharArray();        int[] res = new int[chs.length + 1];        int n = chs.length + 1;        for (int i = 0; i < n; i ++) {            res[i] = i + 1;        }        int start = 0;        for (int i = 0; i < chs.length; i++) {            if (chs[i] == 'D') {                start = i;                while (i < chs.length && chs[i] == 'D') {                    i ++;                }                reverse(res, start, i);            }        }        return res;    }        private void reverse(int[] nums, int s, int e) {        while (s < e) {            int temp = nums[s];            nums[s] = nums[e];            nums[e] = temp;            s ++;            e --;        }    }}