给定只含 "I"(增大)或 "D"(减小)的字符串 S ,令 N = S.length

返回 [0, 1, ..., N] 的任意排列 A 使得对于所有 i = 0, ..., N-1,都有:

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  • 如果 S[i] == "I",那么 A[i] < A[i+1]
  • 如果 S[i] == "D",那么 A[i] > A[i+1]

示例 1:

输出:"IDID"
输出:[0,4,1,3,2]

示例 2:

输出:"III"
输出:[0,1,2,3]

示例 3:

输出:"DDI"
输出:[3,2,0,1]

提示:

  1. 1 <= S.length <= 1000
  2. S 只包含字符 "I""D"
class Solution942 {
    public int[] diStringMatch(String S) {
        int[] nums = new int[S.length() + 1];
        char[] chs = S.toCharArray();

        int n = 0;
        for (char c : chs) {
            if (c == 'D')
                n++;
        }

        int m = n + 1;
        nums[0] = n--;
        for (int i = 0; i < chs.length; i++) {
            if (chs[i] == 'I')
                nums[i + 1] = m++;
            else
                nums[i + 1] = n--;
        }

        return nums;
    }
}
       
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