【牛客题霸题解】反转字符串

解法一:
开辟一个和str长度大小相同的一个字符串ans，把传入的str倒序赋值到ans字符串上， 时间复杂度,额外空间复杂度
c++

class Solution {
public:
    string solve(string str) {
        string ans = str;
        int len = str.length();
        for(int i = 0 ; i < len ;i++)
        {
                ans[i] = str[len-1-i];
        }
        return ans;
    }
};

java

import java.util.*;
public class Solution {
    public String solve (String str) {
        char[] ans = str.toCharArray();
        int len = str.length();
        for(int i = 0 ; i < len ;i++)
        {
                ans[i] = str.charAt(len-1-i);
        }
        return new String(ans);
    }
}

python

class Solution:
    def solve(self , str ):
        ans=""
        Len = len(str)
        for i in range(0,Len):
            ans+=str[Len-1-i]
        return ans

解法二:
原地交换，,注意：交换进行的次数是次
时间复杂度,额外空间复杂度

class Solution {
public:
    string solve(string str) {
        int len = str.length();
        for(int i = 0 ; i < len/2 ;i++)
        {
                swap(str[i],str[len-1-i]);
        }
        return str;
    }
};

java

import java.util.*;
public class Solution {
    public String solve (String str) {
        char[] cstr = str.toCharArray();
        int len = str.length();
        for(int i = 0 ; i < len/2 ;i++)
        {
                char t = cstr[i];
                cstr[i] = cstr[len-1-i];
                cstr[len-1-i]=t;
        }
        return new String(cstr);
    }
}

解法三:
直接调用库函数
c++

class Solution {
public:
    string solve(string str) {
       reverse(str.begin(),str.end());
       return str;
    }
};

java

import java.util.*;
public class Solution {
    public String solve (String str) {
        StringBuffer sb =new StringBuffer(str);//此方法针对的是io流，不能针对字符串。
        return sb.reverse().toString();
    }
}

python

class Solution:
    def solve(self , str ):
        return str[::-1]