LeetCode Find Permutation

原题链接在这里：https://leetcode.com/problems/find-permutation/description/

题目：

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

题解：

先生成sorted array. 若出现连续的D时就把连续D开始和结尾对应的这段reverse. 

Time Complexity: O(n). n = s.length().

Space: O(1). regardless res.

AC Java:

class Solution {
    public int[] findPermutation(String s) {
        int len = s.length();
        int [] res = new int[len+1];
        for(int i = 0; i<len+1; i++){
            res[i] = i+1;
        }
        
        for(int i = 0; i<len; i++){
            if(s.charAt(i) == 'D'){
                int mark = i;
                while(i<len && s.charAt(i)=='D'){
                    i++;
                }
                reverse(res, mark, i);
            }
        }
        
        return res;
    }
    
    private void reverse(int [] res, int i, int j){
        while(i<j){
            swap(res, i++, j--);
        }
    }
    
    private void swap(int [] res, int i, int j){
        int temp = res[i];
        res[i] = res[j];
        res[j] = temp;
    }
}