[Leetcode]942. DI String Match

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本文详细解析了DI字符串匹配算法，这是一种针对特定字符串'I'和'D'的排列问题的高效解决方案。文章通过实例展示了如何根据输入字符串生成符合升序'I'和降序'D'条件的数字排列，并提供了O(N)时间复杂度的算法实现。

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Easy
Related Topic: Math

Problem

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”.

O(N)

class Solution:
    def diStringMatch(self, S):
        """
        :type S: str
        :rtype: List[int]
        """
        res = []
        l = len(S)
        h = 0
        t = l
        for g in S:
            if g == "I":
                res.append(h)
                h += 1
            if g == "D":
                res.append(t)
                t -= 1
        res.append(h)
        return res