【394】Decode String

题目：

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".

思路：

定义一个函数string decodeString(const string& s,int& i),用来表示从第该字符串的下标 i 处开始decode；按数字 [ decodeString(s,i) ] 的方式进行迭代，直到得出最终结果。

代码：

class Solution {public:        string decodeString(const string& s,int& i){        string op;                while(i<s.length() && s[i]!=']') {            if(!isdigit(s[i])){                op += s[i];                i++;            }            else {                int n=0;                while(i<s.length() && isdigit(s[i])) {                    n = n*10 + s[i] - '0';                    i++;                }                                i++;                string temp = decodeString(s,i);                i++;                                while(n>0) {                    op+=temp;                    n--;                }                            }                    }        return op;    }        string decodeString(string s) {        int i=0;        return decodeString(s,i);    }};