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"
, thenA[i] < A[i+1]
- If
S[i] == "D"
, thenA[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"
.
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:
- 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 <= 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] = v 14 v -= 1 15 } 16 pre = i + 1 17 } 18 } 19 for j in (pre...(n - 1)).reversed() 20 { 21 ret[j] = v 22 v -= 1 23 } 24 return ret 25 } 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 }