# 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;
}
}
``````