541. Reverse String II LeetCode

最新推荐文章于 2025-01-10 07:45:24 发布

Edwardwe

最新推荐文章于 2025-01-10 07:45:24 发布

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分类专栏： C++，LeetCode

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本文介绍了一种特殊的字符串反转算法，该算法每隔2k个字符将前k个字符进行反转，并提供了详细的实现过程及代码示例。

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题目：

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:

The string consists of lower English letters only.
Length of the given string and k will in the range [1, 10000]

题目分析：好像是没什么需要分析的吧~~~

class Solution {

public:
    string reverse(string str,int length)//这是简单的逆置字符串的算法
    {
        for (int i = 0; i < length / 2; ++i)  
       {  
        str[length] = str[i];  
        str[i] = str[length - 1 - i];  
        str[length - 1 - i] = str[length];  
       }  
      
       return str; 
    }
    string reverseStr(string s, int k) {
        int len=s.length();
        string temp="";
        int count=0;
        while(len>0)
        {
            if(len>=2*k)//三种情况
            {
                string temp1=s.substr(count,k);
                temp1=reverse(temp1,k);//一半逆置
                temp=temp+temp1;
                count+=k;
                string temp2=s.substr(count,k);
                temp=temp+temp2;//一半直接粘贴
                len-=(2*k);
                count+=k;
                if(len==0) return temp;
            }
            else if(len<2*k&&len>=k)
            {
                string temp1=s.substr(count,k);
                temp1=reverse(temp1,k);
                temp=temp+temp1;
                count+=k;
                string temp2=s.substr(count);
                temp=temp+temp2;
                return temp;
            }
            else if(len<k)
            {
                string temp1=s.substr(count,len);
                temp1=reverse(temp1,len);
                temp=temp+temp1;
                return temp;
            }
        }
    }
};