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

一刷
题解：
For example, given IDIIDD we start with sorted sequence 1234567
Then for each k continuous D starting at index i we need to reverse [i, i+k] portion of the sorted sequence.

IDIIDD
1234567 // sorted
1324765 // answer

public int[] findPermutation(String s) {
        int n = s.length(), arr[] = new int[n + 1]; 
        for (int i = 0; i <= n; i++) arr[i] = i + 1; // sorted
        for (int h = 0; h < n; h++) {
            if (s.charAt(h) == 'D') {
                int l = h;
                while (h < n && s.charAt(h) == 'D') h++;
                reverse(arr, l, h); 
            }   
        }   
        return arr;
    }   

    void reverse(int[] arr, int l, int h) {
        while (l < h) {
            arr[l] ^= arr[h];
            arr[h] ^= arr[l];
            arr[l] ^= arr[h];
            l++; h--;
        }   
    }