力扣 394 字符串解码 栈

1、

2、

3[a]2[bc]

3   num= 3

[    numstack : 3

a cur=a

]  迭代  strstack aaa  cur=aaa stack pop

2 num=2

[  numstack : 2

bc  strstack.push(cur) cur.clear()  strstack aaa

] strstrack aaabcbc

主要思路使用字符串栈来保存结果

3、

class Solution {
public:
    string decodeString(string s) {
        int len = s.size();
        int num = 0;
        stack numstack;
        stack strstack;
        string cur = "";
        string result = "";
        for(int i = 0;i < len;i++)
        {
            if(s[i] >='0' && s[i] <= '9')
            {
                num = 10*num + s[i] - '0';
            }
            else if((s[i] >= 'a' && s[i] <= 'z') || (s[i] >= 'A' && s[i] <= 'Z'))
            {
                cur+= s[i];
            }
            else if(s[i] == '[')
            {
                numstack.push(num);
                num=0;
                strstack.push(cur);
                cur.clear();
            }
            else if(s[i] == ']')
            {              
                for(int j = 0;j < numstack.top();j++)
                {
                    strstack.top()+=cur;
                }
                numstack.pop();
                cur = strstack.top();
                strstack.pop();

            }
        }
        return cur;
    }
};

4、大佬写的递归看起来要简单的多

class Solution {
public:
string analysis(string s, int& index) {
    string res;
    int num = 0;
    string temp;
    while (index < s.size()) {
        if (s[index] >= '0' && s[index] <= '9') {
            num = 10 * num + s[index] - '0';
        }
        else if (s[index] == '[') {
            temp = analysis(s, ++index);//碰到'[',开始递归
            while(num-->0) res += temp;
            num = 0;//num置零
        }
        else if (s[index] == ']') break;//碰到']',结束递归
        else res += s[index];
        index++;
    }
    return res;
}
string decodeString(string s) {
    int index = 0;
    return analysis(s, index);
}
};