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

1.使用自己定义的reverse函数

class Solution {
public:
    string Reverse(string& s,int begin,int end){
        for(int i=begin,j=end-1;i<j;i++,j--){
            swap(s[i],s[j]);
        }
        return s;
    }
    string reverseStr(string s, int k) {
        int length=s.size();
        for(int i=0;i<length;i+=2*k){
            if(i+k<=length)Reverse(s,i,i+k);
            else Reverse(s,i,length);
        }
        return s;
    }
};

注意：

        1.定义自己的reverse函数，需要用&符号，否则无法改变字符串

        2.Reverse函数中i，j的起始位置需要注意

2.使用C++的库函数reverse

class Solution {
public:
    string reverseStr(string s, int k) {
        int length=s.size();
        for(int i=0;i<length;i+=2*k){
            if(i+k<=length)reverse(s.begin()+i,s.begin()+i+k);
            else reverse(s.begin()+i,s.end());
        }
        return s;
    }
};

注意：

        1,C++库的reverse函数的输入

        2.整体思路：如果剩余部分大于等于k，前k个反转，小于k，全部反转