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484. Find Permutation

Problem:

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:

  1. The input string will only contain the character 'D' and 'I'.
  2. The length of input string is a positive integer and will not exceed 10,000

Solutions:

public class Solution {
    public int[] findPermutation(String s) {
        int[] down = new int[s.length() + 1];
        int count = 0;
        for (int i = s.length() - 1; i >= 0; i --) {
            if (s.charAt(i) == 'D') {
                count ++;
            }
            else {
                down[i+1] = count;
                count = 0;
            }
        }
        down[0] = count;
        int next = 1;
        int j = 0;
        while ( j < down.length) {
            if (down[j] != 0) {
                int repeat = down[j];
                down[j] = next + repeat;
                next = down[j] + 1;
                for (int k = 1; k <= repeat; k ++) {
                    down[j + k] = down[j] - k;
                }
                j = j + repeat + 1;
            }
            else {
                down[j++] = next++;
            }
        }
        return down;
    }
}