文章目录
一、题目
541. Reverse String II
Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.
If there are fewer 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 leave the other as original.
Example 1:
Input: s = "abcdefg", k = 2
Output: "bacdfeg"
Example 2:
Input: s = "abcd", k = 2
Output: "bacd"
Constraints:
1 <= s.length <= 104
s consists of only lowercase English letters.
1 <= k <= 104
二、题解
cpp
class Solution {
public:
string reverseStr(string s, int k) {
int n = s.length();
//循环处理2k个单位
for(int i = 0;i < n;i += 2 * k){
if(i + k <= n){
reverse(s.begin() + i,s.begin() + i + k);
}
else reverse(s.begin() + i,s.end());
}
return s;
}
};