484. Find Permutation

Description

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

找规律，time O(n), space O(1)

For example, given IDIIDD we start with sorted sequence 1234567
Then for each k continuous D starting at index i we need to reverse [i, i+k] portion of the sorted sequence.

IDIIDD
1234567 // sorted
1324765 // answer

class Solution {
    public int[] findPermutation(String s) {
        int n = s.length();
        int[] permutation = new int[n + 1];
        
        for (int i = 0; i <= n; ++i) {
            permutation[i] = i + 1;     // sorted
        }
        
        int dStart = 0;
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) != 'D') continue;
            
            int left = i;
            while (i < n && s.charAt(i) == 'D') {
                ++i;
            }
            reverse(permutation, left, i);
        }
        
        return permutation;
    }
    
    public void reverse(int[] nums, int start, int end) {
        while (start < end) {
            int tmp = nums[start];
            nums[start++] = nums[end];
            nums[end--] = tmp;
        }
    }
}