leetcode Letter Combinations of a Phone Number

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递归方法解决。

/**

• Return an array of size *returnSize.

• Note: The returned array must be malloced, assume caller calls free().
  /
  char* letterCombinations(char* digits, int* returnSize) {

  if(strcmp("",digits)==0){*returnSize=0;return “”;}
  char map[9][4]={“00”,“abc”,“def”,“ghi”,“jkl”,“mno”,“pqrs”,“tuv”,“wxyz”};

  char ** a =(char**)malloc(2000*sizeof(char *));

  char * a1 = (char*)malloc(sizeof(char) * 100);
  int ptr=0;
  int outptr=0;

  bool digui(int outptr,char * b)
  {
  // printf("~~~%d %s\n",outptr,b);
  if(outptrstrlen(digits))
  {
  char * lin=(char*)malloc(sizeof(char) * 100);
  strncpy(lin,b,outptr);
  lin[outptr]=’\0’;
  // printf("%s %s ",lin,b);
  a[ptr]=lin;
  ptr++;
  return 1;
  }
  if(digits[outptr]‘2’)
  {
  b[outptr]=‘a’;
  digui(outptr+1,b);
  b[outptr]=‘b’;
  digui(outptr+1,b);
  b[outptr]=‘c’;
  digui(outptr+1,b);
  }
  else if(digits[outptr]‘3’)
  {
  b[outptr]=‘d’;
  digui(outptr+1,b);
  b[outptr]=‘e’;
  digui(outptr+1,b);
  b[outptr]=‘f’;
  digui(outptr+1,b);
  }
  else if(digits[outptr]‘4’)
  {
  b[outptr]=‘g’;
  digui(outptr+1,b);
  b[outptr]=‘h’;
  digui(outptr+1,b);
  b[outptr]=‘i’;
  digui(outptr+1,b);
  }
  else if(digits[outptr]‘5’)
  {
  b[outptr]=‘j’;
  digui(outptr+1,b);
  b[outptr]=‘k’;
  digui(outptr+1,b);
  b[outptr]=‘l’;
  digui(outptr+1,b);
  }
  else if(digits[outptr]‘6’)
  {
  b[outptr]=‘m’;
  digui(outptr+1,b);
  b[outptr]=‘n’;
  digui(outptr+1,b);
  b[outptr]=‘o’;
  digui(outptr+1,b);
  }
  else if(digits[outptr]‘7’)
  {
  b[outptr]=‘p’;
  digui(outptr+1,b);
  b[outptr]=‘q’;
  digui(outptr+1,b);
  b[outptr]=‘r’;
  digui(outptr+1,b);
  b[outptr]=‘s’;
  digui(outptr+1,b);
  }
  else if(digits[outptr]‘8’)
  {
  b[outptr]=‘t’;
  digui(outptr+1,b);
  b[outptr]=‘u’;
  digui(outptr+1,b);
  b[outptr]=‘v’;
  digui(outptr+1,b);
  }
  else if(digits[outptr]==‘9’)
  {
  b[outptr]=‘w’;
  digui(outptr+1,b);
  b[outptr]=‘x’;
  digui(outptr+1,b);
  b[outptr]=‘y’;
  digui(outptr+1,b);
  b[outptr]=‘z’;
  digui(outptr+1,b);
  }
  //,
  return 0;
  }
  digui(outptr,a1);

  *returnSize=ptr;
  return a;
  }