LeetCode 17. Letter Combinations of a Phone Number ( C ) – Medium (遞迴法)
Posted On 2021 年 6 月 12 日
傳統手機透過數字鍵來打字,一個數字鍵通常代表了多個英文字母,例如 2 代表 a , b , c ,3 代表 d , e , f 等,連續按多個鍵能產生英文字母的組合,此題為給定數字,輸出所有可能的英文字母組合。
原文題目如下
Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. Return the answer in any order.
A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
Example 1:
Input: digits = "23"
Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]
Example 2:
Input: digits = ""
Output: []
Example 3:
Input: digits = "2"
Output: ["a","b","c"]
Constraints:
0 <= digits.length <= 4
digits[i] is a digit in the range ['2', '9'].
Accepted
861,326
Submissions
1,706,491
解題方式為透過遞迴的方式,透過 for 迴圈把每個數字可能的英文字母遍歷出來,分別遞迴。填完字母後,補上結束字元 ‘ \0 ‘ 轉存回要回傳的空間內。
下面是我的程式碼
char map[10][5]={{},{},{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'},{'m','n','o'},{'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}}; int reindex = 0; char **re; char *DPtr; void Rnum( int index, char *tempstr){ // index use for input string index and tempstr index if(DPtr[index] == '\0'){ tempstr[index] = '\0'; re[reindex] = (char*)malloc(sizeof(char)*(strlen(DPtr)+1)); strcpy(re[reindex],tempstr); reindex++; } else{ for(int i = 0;i<strlen(map[DPtr[index]-'0']);i++){ tempstr[index] = map[DPtr[index]-'0'][i]; Rnum(index+1,tempstr); } } } char ** letterCombinations(char * digits, int* returnSize){ DPtr = digits; reindex = 0; if(strlen(digits) != 0){ (*returnSize) = 1; for(int i = 0 ; i < strlen(digits) ; i++ ){ (*returnSize) *= strlen(map[digits[i]-'0']); } } else{ (*returnSize) = 0; } re = (char**)malloc(sizeof(char*)*(*returnSize)); char tempstr[5] = {0}; if((*returnSize)>0) Rnum( 0, tempstr); return re; }
下面是時間與空間之消耗
Runtime: 0 ms, faster than 100 % of C online submissions for Letter Combinations of a Phone Number.
Memory Usage: 5.7 MB, less than 98.79 % of C online submissions for Letter Combinations of a Phone Number.