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:

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

Tips: Greedy, O(n)

设置一个min。
1. 如果是I，就res[i++] = min++;
2. 如果是D，就遍历到不是D的地方，然后从大到小加数字。

Code:

public class Solution {
    public int[] findPermutation(String s) {
        s = s + ".";
        int[] res = new int[s.length()];
        int min = 1, i = 0;
        while (i < res.length) {
            while (s.charAt(i) == 'I') res[i++] = min++;
            int j = i;
            while (s.charAt(j) == 'D') j++;
            for (int k = j; k >= i; k--) res[k] = min++;
            i = j + 1;
        }
        return res;
    }
}