[C语言][LeetCode][394]Decode String

题目

Decode String
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”.

标签

Stack、Depth-first Search

难度

中等

分析

题目意思是提供一个字符串，然后按照例子的格式，解密这个字符串。
实现思路是利用递归的思想，遇到左括号，进入递归操作，遇到右括号，则返回字符串，具体见代码实现。

C代码实现

char* dfs(char* s, int *k) {    int i=0, cnt=0, len=0;    char * temp = NULL;    char * res = (char *)malloc(sizeof(char));    memset(res, 0, sizeof(char));    while(*k < strlen(s))    {        //printf("Input char : %2c\n", s[*k]);        if( ('0'<=s[*k]) && ('9'>=s[*k]) )        {                        cnt = cnt*10 + (s[*k]-'0');            *k = *k + 1;        }        else if(s[*k] == '[')        {            *k = *k + 1;            temp = dfs(s, k);                        len = strlen(temp);                        //printf("\nReturn String : (len :%3d) %s\n", len, temp);            if(strlen(res)<strlen(temp))                res = (char *)realloc(res, cnt*len*sizeof(char)+1);            else                 res = (char *)realloc(res, cnt*len*sizeof(char)+strlen(res)+1);                if(res == NULL)                {                    printf("realloc is NULL, fail\n");                }                else                {                    for(i=0; i<cnt; i++)                    {                        strncat(res, temp, len);                        //printf("Splice String : (time:%3d) %s\n", i+1, res);                    }                    cnt = 0;                }        }        else if(s[*k] == ']')        {            *k = *k + 1;            return res;        }        else         {            strncat(res, &s[*k], 1);            if(strlen(res)>1)                res = (char *)realloc(res, sizeof(char)+strlen(res));            else                res = (char *)realloc(res, sizeof(char));            *k = *k + 1;        }    }    return res;}char* decodeString(char* s) {    int k = 0;    char * result = dfs(s, &k);    result[strlen(result)] = '\0';    //printf("\nGet Decode string len : %d\n", strlen(result));    return result;}