Letter Combinations of a Phone Number(回溯，dfs）

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

先看题目，题目要求按下某些按键后，返回所有的字母的排列组合，注意这题的输入只接受2-9.

po个代码

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
static const char* table[10] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
void dfs(char **table,int index,char *digits,char *res,int length,int *used,int *count,char **ans)
{
    if(digits[index]=='\0')
    {
        char *temp=(char *)calloc(length,sizeof(char ));
        for(int i=0;res[i]!='\0';i++)
        {
            temp[i]=res[i];
        }
        *(ans+*count)=temp;
        *count=*count+1;
        return ;
    }
    else
    {
        for(int i=index;i<length;i++)
        {
            for(int j=0;table[digits[i]-'0'][j]!='\0';j++)
            {
                if(used[i]==0)
                {
                    used[i]=1;
                    res[index]=table[digits[i]-'0'][j];
                    dfs(table,index+1,digits,res,length,used,count,ans);  
                    used[i]=0;
                }
            }
        }
    }
}
char** letterCombinations(char* digits, int* returnSize) 
{
    if(digits[0]=='\0')
    {
        return "";
    }
    for(int i=0;digits[i]!='\0';i++)
    {
        if(digits[i]<'2'||digits[i]>'9')
        {
            return "";
        }
    }
    int length=strlen(digits);
    char *res=(char *)calloc(length,sizeof(char ));
    int *used=(int *)calloc(length,sizeof(int ));
    int *count=(int *)malloc(sizeof(int ));
    *count=0;
    char **ans=(char **)calloc(1024,sizeof(char *));
    dfs(table,0,digits,res,length,used,count,ans);
    *returnSize=*count;
    return ans;
}

这题先定义一个常量用来记录按键中的字母，然后建立解答树，dfs

可以发现这题的框架和前面一题一样

void dfs()

{

if(now==something)

{

return ;

}

for(.....)

{

....

}

}

这题主要是c语言写的地址传来传去看上去复杂一点