Rust 解法, 执行用时: 2ms, 内存消耗: 228KB, 提交时间: 2021-03-01

struct Solution{

}

impl Solution {
    fn new() -> Self {
        Solution{}
    }

    /**
    * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
    * 
        * @param ReverseSentence string字符串 
        * @return string字符串
    */
    pub fn ReverseSentence(&self, ReverseSentence: String) -> String {
        // write code here
        ReverseSentence.split(' ').into_iter().rev().collect::<Vec<_>>().join(" ") 
    }
}

C++ 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2021-04-30

class Solution {
public:
    string ReverseSentence(string str) {
        string temp = "";
        string ans = "";
        int len = str.length();
        for(int i=len-1;i>=0;i--)
        {
            if(str[i]!=' ')
            {
                temp=str[i]+temp;
            }else
            {
                ans += temp + ' ';
                temp = "";
            }
        }
        if(temp.length()!=0)
        {
            ans += temp;
        }
        return ans;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2021-04-21

class Solution {
public:
    string ReverseSentence(string str) {
        int left = 0, right = str.length()-1;
        while(right>=0 && str[right]==' ') right--;
        if(right == -1) return str;
        
        int len = right+1;
        str.resize(len);
        while(left < right) swap(str[ left++ ], str[ right-- ]);
        
        int i = left = 0;
        while(i < len)
        {
            if(str[i] == ' '){
                right = i-1;
                while(left < right) swap(str[ left++ ], str[ right-- ]);
                left = ++i;
                continue;
            }
            i++;
        }
        right = i-1;
        while(left < right) swap(str[ left++ ], str[ right-- ]);
        
        return str;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2021-04-05

class Solution {
public:
    string ReverseSentence(string str) {
        int len = 0, num = str.size(), pos = 0;
        vector<string> v;
        string st;
        //特殊情况1:字符串为空，则原样返回
        if(str.empty()) return str;
        //特殊情况2：字符串全是空格，则原样返回
        while(str[len] == ' ')
            len++;
        if(len == str.size())
            return str;
        //普通判断,将字符串以空格为间隔拆分为多个字符串，之后再进行拼接
        while(num > 0)
        {
            while(str[pos+len] != ' ' && num > 0)
            {
                len++;
                num--;
            }
            v.push_back(str.substr(pos,len));
            num--;
            len++;
            pos += len;
            len = 0;
        }
        for(int i = v.size()-1; i >= 0; i--)
        {
            st.append(v[i]);
            st.append(" ");
        }
        st.erase(st.size()-1,1);
        return st;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2021-04-04

class Solution {
public:
    string ReverseSentence(string str) {
        string res;
        string temp;
        for (auto c: str) {
            if (c == ' ') {
                res = temp + ' ' + res;
                temp = "";
            }
            else {
                temp += c;
            }
        }
        res = temp + ' ' + res;
        res = res.substr(0, res.size() - 1);
        return res;
    }
};