题目
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.
分析
手机上的拼音输入法，每个数字键代表多个字母。
根据输入的数字串，返回可能的字符串
下面的C代码已通过。
先初始化二维字符数组，然后如果出现0或者1，或者数字串为空，就直接返回空。然后再依次递归判断接下来的字符，根据不同的数字表示不同的字符，之后在找到最后一个字符的时候，将该字符串存入结果字符串数组中。
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
void combination(char *digits,int digitsLength,char ** ans,int *returnSize,char *temp,int tempLength)
{
if(digits[digitsLength]==NULL)
{
char * ansTemp=(char *)malloc(sizeof(char)*1000);
int i=0;
for(i=0;i<tempLength;i++)
ansTemp[i]=temp[i];
ansTemp[i]='\0';
ans[*returnSize]=ansTemp;
*returnSize=*returnSize+1;
//printf("%s %d\n",ansTemp,*returnSize);
}
else if(digits[digitsLength]=='2')
{
digitsLength++;
temp[tempLength]='a';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='b';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='c';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
tempLength++;
}
else if(digits[digitsLength]=='3')
{
digitsLength++;
temp[tempLength]='d';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='e';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='f';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
tempLength++;
}
else if(digits[digitsLength]=='4')
{
digitsLength++;
temp[tempLength]='g';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='h';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='i';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
tempLength++;
}
else if(digits[digitsLength]=='5')
{
digitsLength++;
temp[tempLength]='j';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='k';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='l';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
tempLength++;
}
else if(digits[digitsLength]=='6')
{
digitsLength++;
temp[tempLength]='m';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='n';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='o';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
tempLength++;
}
else if(digits[digitsLength]=='7')
{
digitsLength++;
temp[tempLength]='p';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='q';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='r';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='s';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
tempLength++;
}
else if(digits[digitsLength]=='8')
{
digitsLength++;
temp[tempLength]='t';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='u';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='v';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
tempLength++;
}
else if(digits[digitsLength]=='9')
{
digitsLength++;
temp[tempLength]='w';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='x';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='y';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
temp[tempLength]='z';
combination(digits,digitsLength,ans,returnSize,temp,tempLength+1);
tempLength++;
}
}
char** letterCombinations(char* digits, int* returnSize) {
int i=0;
char **ans=(char **)malloc(sizeof(char *)*1000000);
*returnSize=0;
char *temp=(char *)malloc(sizeof(char)*1000);
if(digits[i]=='\0')
return NULL;
while(digits[i]!='\0')
{
if(digits[i]=='0'||digits[i]=='1')
return NULL;
i++;
}
combination(digits,0,ans,returnSize,temp,0);
return ans;
}
当然还有更简单的方法，先处理前两个数字尽可能多的字符串，之后依次根据后面的数字再组成更多的字符串，可能需要重复赋值，但是相对的内存空间消耗较少。而我的方法多次递归，消耗大量栈空间。
