1221 Split a String in Balanced Strings 分割平衡字符串
Description:
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:
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".
Example 4:
Input: s = "RLRRRLLRLL"
Output: 2
Explanation: s can be split into "RL", "RRRLLRLL", since each substring contains an equal number of 'L' and 'R'
Constraints:
1 <= s.length <= 1000
s[i] = 'L' or 'R'
题目描述:
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 :
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'
思路:
相当于 R和 L从头开始匹配, 相同的压入栈, 不同的弹出栈, 只有栈中的完全匹配(栈空, 即L和 R的数量相等), 结果➕1
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int balancedStringSplit(string s)
{
int result = 0, count = 0;
for (auto c : s)
{
if (c == 'L') ++count;
else --count;
if (!count) ++result;
}
return result;
}
};
Java:
class Solution {
public int balancedStringSplit(String s) {
int result = 0, count = 0;
for (char c : s.toCharArray()) {
if (c == 'L') ++count;
else --count;
if (count == 0) ++result;
}
return result;
}
}
Python:
class Solution:
def balancedStringSplit(self, s: str) -> int:
result, count = 0, 0
for i in s:
count += 1 if i == 'L' else -1
if not count:
result += 1
return result