一道stack的题目

原题如下
Given an encoded string, return it’s decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won’t be input like 3a or 2[4].

Examples:

s = “3[a]2[bc]”, return “aaabcbc”.
s = “3[a2[c]]”, return “accaccacc”.
s = “2[abc]3[cd]ef”, return “abcabccdcdcdef”.

看到括号匹配首先想到用栈来解决问题。
所以遍历字符串当发现‘[’时将当前i压入栈，发现‘]’时，将栈顶弹出并处理。
其中要注意处理完后i的取值。
代码如下：

string decodeString(string s) {    stack<int>p;    for(int i=0;i<s.size();i++){        if(s[i]=='[') p.push(i);        if(s[i]==']'){            int k=p.top();            p.pop();            string s1=s.substr(0,k);            string s2=s.substr(i+1);            string s3=s.substr(k+1,i-k-1);            int num=0;            int total=0;            int len=s1.size()-1;            while(len>=0&&s[len]<='9'&&s[len]>='0'){                len--;            }            for(int j=len+1;j<s1.size();j++){                num*=10;                num+=s1[j]-'0';            }            if(len==-1) s1="";            else            s1=s1.substr(0,len+1);            s=s1;            i=s1.length()+num*s3.length()-1;            while(num--) s+=s3;            s+=s2;         }     }        return s;    }