[LeetCode] 541.Reverse String II
- 题目描述
- 解题思路
- 实验代码
题目描述
Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.
Example:
Input: s = “abcdefg”, k = 2
Output: “bacdfeg”Restrictions:
1. The string consists of lower English letters only.
2. Length of the given string and k will in the range [1, 10000]
解题思路
和上次的字符串反转一样,这道题我还是用到了reverse()函数,而难点在于理解题目意思,需要做一个判断,由于2*k的不确定性,想清楚就能很容易解决。
实验代码
class Solution {
public:
string reverseStr(string s, int k) {
int l = s.length();
for (int i = 0; i < l; i += 2*k)
reverse(s.begin()+i, min(s.begin()+i+k, s.end()));
return s;
}
};
本文介绍了解决LeetCode 541题目的方法,该题目要求每2k个字符中反转前k个字符。通过使用C++实现,文章详细解析了如何针对不同情况正确地进行字符串的反转。
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