942. DI String Match

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]

解题要点：

遍历字符串（转换成数组），遇D，从字符串长度开始加；遇I，从0开始加。最后加入从零开始的那个值，返回数组。

class Solution {
    public int[] diStringMatch(String S) {
        int N = S.length();
        int[] res = new int[N + 1];
        if(N == 0) return res;
        int start = 0, count = 0;
        for(char c : S.toCharArray()){
            if(c == 'I') res[count++] = start++;
            if(c == 'D') res[count++] = N--;
        }
        res[count] = start;
        return res;
    }
}