[Swift]LeetCode1221. 分割平衡字符串 | Split a String in Balanced Strings

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Balanced strings are those who have equal quantity of 'L' and 'R' characters.

Given a balanced string s split it in the maximum amount of balanced strings.

Return the maximum amount of splitted balanced strings.

Example 1:

Input: s = "RLRRLLRLRL"

Output: 4

Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.

Example 2:

Input: s = "RLLLLRRRLR"

Output: 3

Explanation: s can be split into "RL", "LLLRRR", "LR", each substring contains same number of 'L' and 'R'.

Example 3:

Input: s = "LLLLRRRR"

Output: 1

Explanation: s can be split into "LLLLRRRR".

Constraints:

1 <= s.length <= 1000

s[i] = 'L' or 'R'

在一个「平衡字符串」中，'L' 和 'R' 字符的数量是相同的。

给出一个平衡字符串 s，请你将它分割成尽可能多的平衡字符串。

返回可以通过分割得到的平衡字符串的最大数量。

示例 1：

输入：s = "RLRRLLRLRL"

输出：4

解释：s 可以分割为 "RL", "RRLL", "RL", "RL", 每个子字符串中都包含相同数量的 'L' 和 'R'。

示例 2：

输入：s = "RLLLLRRRLR"

输出：3

解释：s 可以分割为 "RL", "LLLRRR", "LR", 每个子字符串中都包含相同数量的 'L' 和 'R'。

示例 3：

输入：s = "LLLLRRRR"

输出：1

解释：s 只能保持原样 "LLLLRRRR".

提示：

1 <= s.length <= 1000

s[i] = 'L' 或 'R'

0ms

 1 class Solution {
 2     func balancedStringSplit(_ s: String) -> Int {
 3         if s.isEmpty { return 0 }
 4         
 5         var r = 0
 6         var l = 0
 7         var count = 0
 8         var arrayString = Array(s)
 9         for var i in 0..<arrayString.count {
10             if arrayString[i] == "L" {
11                 l += 1
12             } else {
13                 r += 1
14             }
15             
16             if r == l {
17                 count += 1
18             }
19         }
20         return count
21     }
22 }

4ms

 1 class Solution {
 2     func balancedStringSplit(_ s: String) -> Int {
 3         var balance = 0
 4         var count = 0 
 5         
 6         for cha in s {
 7             if cha == "R" {
 8                 balance += 1
 9             } else {
10                 balance -= 1
11             }
12             
13             if balance == 0 {
14                 count += 1
15             }
16         }
17         
18         return count
19     }
20 }

8ms

 1 class Solution {
 2     func balancedStringSplit(_ s: String) -> Int {
 3         var b = 0
 4         var ans = 0
 5         for c in s {
 6             if c == "L" {
 7                 b -= 1
 8             } else {
 9                 b += 1
10             }
11             if b == 0 {
12                 ans += 1
13             }
14         }
15         return ans
16     }
17 }

12ms

 1 class Solution {
 2     func balancedStringSplit(_ s: String) -> Int {
 3         var counter = 0
 4         var balancedStrings = [String]()
 5         var balancedString = ""
 6         for char in s {
 7             var val = char == "R" ? 1 : -1
 8             counter += val
 9             
10             if counter == 0 {
11                 balancedStrings.append(balancedString)
12                 balancedString = ""
13             } else {
14                 balancedString += String(char)
15             }
16         }
17         return balancedStrings.count
18     }
19 }