[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

class Solution {
    func balancedStringSplit(_ s: String) -> Int {
        if s.isEmpty { return 0 }
        
        var r = 0
        var l = 0
        var count = 0
        var arrayString = Array(s)
        for var i in 0..<arrayString.count {
            if arrayString[i] == "L" {
                l += 1
            } else {
                r += 1
            }
            
            if r == l {
                count += 1
            }
        }
        return count
    }
}

---

4ms

class Solution {
    func balancedStringSplit(_ s: String) -> Int {
        var balance = 0
        var count = 0 
        
        for cha in s {
            if cha == "R" {
                balance += 1
            } else {
                balance -= 1
            }
            
            if balance == 0 {
                count += 1
            }
        }
        
        return count
    }
}

---

8ms

class Solution {
    func balancedStringSplit(_ s: String) -> Int {
        var b = 0
        var ans = 0
        for c in s {
            if c == "L" {
                b -= 1
            } else {
                b += 1
            }
            if b == 0 {
                ans += 1
            }
        }
        return ans
    }
}

---

12ms

class Solution {
    func balancedStringSplit(_ s: String) -> Int {
        var counter = 0
        var balancedStrings = [String]()
        var balancedString = ""
        for char in s {
            var val = char == "R" ? 1 : -1
            counter += val
            
            if counter == 0 {
                balancedStrings.append(balancedString)
                balancedString = ""
            } else {
                balancedString += String(char)
            }
        }
        return balancedStrings.count
    }
}