942. DI String Match

QUESTION:

Given a string S that only contains “I” (increase) or “D” (decrease), let N = S.length.

Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1:

  • If S[i] == "I", then A[i] < A[i+1]
  • If S[i] == "D", then A[i] > A[i+1]

Example 1:

Input: "IDID"
Output: [0,4,1,3,2]

Example 2:

Input: "III"
Output: [0,1,2,3]

Example 3:

Input: "DDI"
Output: [3,2,0,1]

Note:

  1. 1 <= S.length <= 10000
  2. S only contains characters "I" or "D".

EXPLANATION:

从例子中的一直升序可以看到，

1，2，3都是从小到大往上排的

而降序也是如此

那我们就可以推断出：升序需要从小的那边开始，而降序需要从另外一边开始。

SOLUTION:

class Solution {
    public int[] diStringMatch(String S) {
        int start = 0;
        int end = S.length();
        int[] result = new int[S.length()+1];
        for(int i = 0;i<S.length();i++){
            char c = S.charAt(i);
            switch (c){
                case 'I':
                    result[i] = start;
                    start++;
                    break;
                case 'D':
                    result[i] = end;
                    end--;
                    break;
            }
        }
        result[result.length-1] = S.charAt(S.length()-1)=='I'?start:end;
        return result;
    }
}
--> return <--