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:
class Solution {
// IDI
// 1 3 2 4
public int[] findPermutation(String s) {
int n = s.length() + 1;
int[] res = new int[n];
for (int i = 0; i < n; i ++) {
res[i] = i + 1;
}
int i = 0;
for (int j = 0; j < s.length(); j ++) {
while (j < s.length() && s.charAt(j) == 'D') {
j ++;
}
reverse(res, i, j);
i = j + 1;
// System.out.println(Arrays.toString(res));
}
return res;
}
private void reverse(int[] arr, int i, int j) {
while (i < j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i ++;
j --;
}
}
}