1221. Split a String in Balanced Strings
2021/9/7 23:10:33
本文主要是介绍1221. Split a String in Balanced Strings,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
题目:
Balanced strings are those that have an 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 split 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".
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]
is either'L'
or'R'
.s
is a balanced string.
思路:
easy题,当一个字符串内L和R的数量一样多时即为平衡字符串。基础思路是遍历字符串,用三个计数器,如果遇到L则 l加一;如果遇到R则 r加一;当l和r相等时,说明前面出现了一个平衡字符串,则count加一,然后把l和r清空,直到遍历完字符串,返回count即可。
代码:
class Solution {
public:
int balancedStringSplit(string s) {
int r = 0, l = 0, count = 0;
for (auto a : s) {
if (a == 'R')
r++;
else if (a == 'L')
l++;
if (l != 0 && l == r) {
count++;
l = r = 0;
}
}
return count;
}
};
这篇关于1221. Split a String in Balanced Strings的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!
- 2024-11-23增量更新怎么做?-icode9专业技术文章分享
- 2024-11-23压缩包加密方案有哪些?-icode9专业技术文章分享
- 2024-11-23用shell怎么写一个开机时自动同步远程仓库的代码?-icode9专业技术文章分享
- 2024-11-23webman可以同步自己的仓库吗?-icode9专业技术文章分享
- 2024-11-23在 Webman 中怎么判断是否有某命令进程正在运行?-icode9专业技术文章分享
- 2024-11-23如何重置new Swiper?-icode9专业技术文章分享
- 2024-11-23oss直传有什么好处?-icode9专业技术文章分享
- 2024-11-23如何将oss直传封装成一个组件在其他页面调用时都可以使用?-icode9专业技术文章分享
- 2024-11-23怎么使用laravel 11在代码里获取路由列表?-icode9专业技术文章分享
- 2024-11-22怎么实现ansible playbook 备份代码中命名包含时间戳功能?-icode9专业技术文章分享