C++ 解法, 执行用时: 2ms, 内存消耗: 404KB, 提交时间: 2021-11-13

class Solution {
public:
    /**
     * 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 s string字符串 
     * @return string字符串
     */
    string decodeString(string s) {
        // write code here
        string res = "";
        stack<int> nums;
        stack<string> strs;
        int num = 0;
        int len = s.size();
        for(int i = 0; i< len; ++i){
            if(s[i] >= '0' && s[i] <= '9'){
                num = num * 10 + s[i] - '0';
            }
            else if((s[i] >= 'a' && s[i] <= 'z') || (s[i] >= 'A' && s[i] <= 'Z')){
                res = res + s[i];
            }
            else if(s[i] == '['){ 
                nums.push(num);
                num = 0; 
                strs.push(res);
                res = ""; 
            }
            else{
                int multi = nums.top();
                nums.pop();
                for(int j = 0; j < multi; ++j)
                    strs.top() += res;
                res = strs.top(); 
                strs.pop();
            }
        }
        return res;
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 388KB, 提交时间: 2022-03-20

class Solution {
public:
    /**
     * 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 s string字符串 
     * @return string字符串
     */
    string decodeString(string s) {
        // write code here
        std::string newStr;
        std::string str = s;
        while (decode(str, newStr)) {
            str = newStr;
        }
        
        return newStr;
    }
    
    bool decode(std::string s, std::string& newStr) {
        std::string str;
        std::string subStr;
        int pos1 = -1;
        int pos2 = -1;
        int numPos = -1;
        bool find = false;
        for (int i = s.size() - 1; i >= 0; --i) {
            if (find) {
                if (s[i] >= '0' && s[i] <= '9') {
                    numPos = i;
                    continue;
                }
                
                break;
            }
            if (s[i] == ']') {
                pos2 = i;
            } else if (s[i] == '[') {
                pos1 = i;
                find = true;
            } else if (s[i] >= '0' && s[i] <= '9') {
                numPos = i;
            }
        }
        
        if (pos1 != -1 && pos2 != -1) {
            int cc = pos2 - pos1 -1; //字符个数
            std::string sss = s.substr(numPos);
            
            int dot = std::stoi(s.substr(numPos));
            
            for (int i = 0; i < dot; ++i) {
                subStr += s.substr(pos1 + 1, cc);
            }
            
            s.erase(s.begin() + numPos, s.begin() + pos2 + 1);
            
            for (int j = 0; j < subStr.size(); ++j) {
                s.insert(s.begin() + numPos + j, subStr[j]);
            }
            
            newStr = s;
            
            return true;
        }
        
        if (newStr.empty()) {
            newStr = s;
        }
        
        return false;
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 392KB, 提交时间: 2022-07-30

class Solution {
public:
    /**
     * 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 s string字符串 
     * @return string字符串
     */
    string decodeString(string s) {
        // write code here
        string str = "";
        int mutli = 0;
        string res = "";
        stack<int>st;
        stack<string>alp;
        for(char c:s){
            if(isdigit(c))mutli = mutli*10 + (c - '0');
            else if(isalpha(c))res += c;
            else if(c == '['){
                st.push(mutli);
                alp.push(res);
                mutli = 0;
                res = "";
            }else if(c == ']'){
                int num = st.top();
                st.pop();
                string tmp = "";
                for(int i = 0;i < num;i++){
                    tmp += res;
                }
                res = alp.top() + tmp;
                alp.pop();
            }
        }
        return res;
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 392KB, 提交时间: 2022-07-02

class Solution {
public:
    string decodeString(string s) {
        string res;
        for (int i = 0; i < s.size();)
        {
            if (!isdigit(s[i])) res += s[i ++ ];
            else
            {
                int j = i;
                while (isdigit(s[j])) j ++ ;
                int t = atoi(s.substr(i, j - i).c_str());
                int k = j + 1, sum = 0;
                while (sum >= 0)
                {
                    if (s[k] == '[') sum ++ ;
                    if (s[k] == ']') sum -- ;
                    k ++ ;
                }
                string str = decodeString(s.substr(j + 1, k - j - 2));
                while (t -- ) res += str;
                i = k;
            }
        }
        return res;
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 392KB, 提交时间: 2022-05-13

class Solution {
public:
    /**
     * 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 s string字符串 
     * @return string字符串
     */
    string ans="";
    stack<int> nums;
    stack<string> st;
    // 考虑测试用例 22[ab10[b]]
    string decodeString(string s) {
        // write code here
        int num=0;
        for(int i=0;i<s.size();++i){
            if(isdigit(s[i])){
                num=num*10+s[i]-'0';
            }
            else if(isalpha(s[i])){
                ans+=s[i];
            }
            else if(s[i]=='['){
                st.push(ans);
                nums.push(num);
                num=0;
                ans="";
            }
            else if(s[i]==']'){
                string tmp="";
                int t=nums.top();
                nums.pop();
                for(int j=0;j<t;++j){
                    tmp+=ans;
                }
                ans=st.top()+tmp;
                st.pop();
            }
        }
        return ans;
    }
};