394. Decode String. 字符串

Given an encoded string, return its 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".

來源：力扣（LeetCode）
連接：https://leetcode-cn.com/problems/decode-string

1.字符串中獲取數值

int getDigits() {
        int ret = 0;
        while (ptr < src.size() && isdigit(src[ptr])) {
            ret = ret * 10 + src[ptr++] - '0';
        }
        return ret;
    }

2.經常使用字符判斷函數
isdigit(cur) //當前字符是否爲數字
isalpha(cur) //當前字符是否爲字母

class Solution {
public:
    string src; 
    int ptr = 0;

    int getDigits() {
        int ret = 0;
        while (ptr < src.size() && isdigit(src[ptr])) {
            ret = ret * 10 + src[ptr++] - '0';
        }
        return ret;
    }

    string getString() {
        if (ptr == src.size() || src[ptr] == ']') {
            return "";
        }

        char cur = src[ptr]; 
        int repTime = 1;
        string ret;

        if (isdigit(cur)) {
            repTime = getDigits(); 
            ++ptr;

            string str = getString(); 
            ++ptr;
            while (repTime--) ret += str; 
        } 
        else 
        if (isalpha(cur)) {
            ret += src[ptr];
            ++ptr;
        }
        
        return ret + getString();
    }

    string decodeString(string s) {
        src = s;
        ptr = 0;
        return getString();
    }
};