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

Solution：

思路：
我们来看一个例子：
D D I I D I
1 2 3 4 5 6 7
*3 *2 *1 4 *6 *5 7
我们不难看出，只有D对应的位置附近的数字才需要变换，而且变换方法就是倒置一下字符串，我们要做的就是通过D的位置来确定需要倒置的子字符串的起始位置和长度即可。通过观察，我们需要记录D的起始位置i，还有D的连续个数k，那么我们只需要在数组中倒置[i, i+k]之间的数字即可.
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

class Solution {
    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) {
            int tmp = arr[l];
            arr[l] = arr[h];
            arr[h] = tmp;
            l++; h--;
        }   
    }
}