[Swift Weekly Contest 111]LeetCode942. 增减字符串匹配 | DI String Match

概述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]

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: [0,4,3,2] 

Example 2:

input: "III"Output: [0,2,3] 

Example 3:

input: "DDI"Output: [3,1]

 Note:

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

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

返回 [0,N] 的任意排列 A 使得对于所有 i = 0,N-1，都有：

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：[0,2]

示例 2：

输出："III"输出：[0,3]

示例 3：

输出："DDI"输出：[3,1]

提示：

1 <= S.length <= 1000 S 只包含字符 "I" 或 "D"。

2096ms

 1 class Solution { 2     func distringMatch(_ S: String) -> [Int] { 3         var n:Int = S.count + 1 4         var ret:[Int] = [Int](repeating: 0,count: n) 5         var v:Int = n - 1 6         var pre:Int = 0 7         for i in 0..<(n - 1) 8         { 9             if S[i] == "D"10             {11                 for j in (pre...i).reversed()12                 {13                     ret[j] = v14                     v -= 115                 }16                 pre = i + 117             }18         }19         for j in (pre...(n - 1)).reversed()20         {21             ret[j] = v22             v -= 123         }24         return ret25     }26 }27 28 extension String {29     //subscript函数可以检索数组中的值30     //直接按照索引方式截取指定索引的字符31     subscript (_ i: Int) -> Character {32         //读取字符33         get {return self[index(startIndex,offsetBy: i)]}34     }35 }

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