942. DI String Match

最新推荐文章于 2023-12-19 16:28:01 发布

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本文深入解析了一种基于字符串I和D的排列算法，该算法能够生成符合特定递增和递减规则的整数序列。通过实例演示了算法的工作原理，包括其核心逻辑和证明过程，同时探讨了可能的解空间和算法的必要条件。

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

题解：可以负责任的告诉大家，这道题虽然列为简单题，但是绝不简单。

下面介绍第一种解法：

class Solution {
    public int[] diStringMatch(String S) {
        int Slength = S.length();
        int decr = Slength;
        int incr = 0;
        int[] a = new int[Slength+1];
        for(int i=0;i<Slength;i++){
           if(S.charAt(i)=='I')
               a[i] = incr++;
            else
               a[i] = decr--;
        }
        a[Slength] = incr;
        return a;
    }
}