LeetCode电话号码的字母组合——C

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number

题目描述

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。
给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例:
输入："23"
输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

题解

大概是这样的

1、暴力破解（好像内存出问题了，提交没通过）
直接按照一个一个数字输入，分配空间把对应的字母赋值就行，可以运行，但提交不行，显示30错误，内存出问题，但是又找不出来。。。。。（看看就好）

#include
#include
#include

char** letterCombinations(char* digits, int* returnSize) {
	int n = strlen(digits);
	if (n == 0) {
		*returnSize = 0;
		return NULL;
	}
	int count = 1;
	for (int i = 0; i < n; i++) {							//计算最终个数 
		if (digits[i] == '7' || digits[i] == '9') {
			count *= 4;
		}
		else {
			count *= 3;
		}
	}
	char** res = (char**)malloc(sizeof(char*) * count);		//分配空间 
	for (int i = 0; i < count; i++) {
		res[i] = (char*)malloc(sizeof(char) * (n+1));
	}
	int k = 0;
	int row = 1;
	for (int i = 0; i < n; i++) {
		switch (digits[i]) {
		case '2':
			for (int j = 0; j < row; j++) {					//按间隔复制 
				strcpy(res[j + row], res[j]);
				strcpy(res[j + 2 * row], res[j]);
			}
			for (int j = 0; j < row; j++) {					//依次增加字母 
				res[j][k] = 'a';
				res[j + row][k] = 'b';
				res[j + 2 * row][k] = 'c';
			}
			k++;
			row *= 3;										//扩充间隔 
			break;
		case '3':
			for (int j = 0; j < row; j++){
				strcpy(res[j + row], res[j]);
				strcpy(res[j + 2 * row], res[j]);
			}
			for (int j = 0; j < row; j++){
				res[j][k] = 'd';
				res[j + row][k] = 'e';
				res[j + 2 * row][k] = 'f';
			}
			k++;
			row *= 3;
			break;
		case '4':
			for (int j = 0; j < row; j++){
				strcpy(res[j + row], res[j]);
				strcpy(res[j + 2 * row], res[j]);
			}
			for (int j = 0; j < row; j++){
				res[j][k] = 'g';
				res[j + row][k] = 'h';
				res[j + 2 * row][k] = 'i';
			}
			k++;
			row *= 3;
			break;
		case '5':
			for (int j = 0; j < row; j++){
				strcpy(res[j + row], res[j]);
				strcpy(res[j + 2 * row], res[j]);
			}
			for (int j = 0; j < row; j++){
				res[j][k] = 'j';
				res[j + row][k] = 'k';
				res[j + 2 * row][k] = 'l';
			}
			k++;
			row *= 3;
			break;
		case '6':
			for (int j = 0; j < row; j++){
				strcpy(res[j + row], res[j]);
				strcpy(res[j + 2 * row], res[j]);
			}
			for (int j = 0; j < row; j++){
				res[j][k] = 'm';
				res[j + row][k] = 'n';
				res[j + 2 * row][k] = 'o';
			}
			k++;
			row *= 3;
			break;
		case '7':
			for (int j = 0; j < row; j++){
				strcpy(res[j + row], res[j]);
				strcpy(res[j + 2 * row], res[j]);
				strcpy(res[j + 3 * row], res[j]);
			}
			for (int j = 0; j < row; j++){
				res[j][k] = 'p';
				res[j + row][k] = 'q';
				res[j + 2 * row][k] = 'r';
				res[j + 3 * row][k] = 's';
			}
			k++;
			row *= 4;
			break;
		case '8':
			for (int j = 0; j < row; j++){
				strcpy(res[j + row], res[j]);
				strcpy(res[j + 2 * row], res[j]);
			}
			for (int j = 0; j < row; j++){
				res[j][k] = 't';
				res[j + row][k] = 'u';
				res[j + 2 * row][k] = 'v';
			}
			k++;
			row *= 3;
			break;
		case '9':
			for (int j = 0; j < row; j++){
				strcpy(res[j + row], res[j]);
				strcpy(res[j + 2 * row], res[j]);
				strcpy(res[j + 3 * row], res[j]);
			}
			for (int j = 0; j < row; j++){
				res[j][k] = 'w';
				res[j + row][k] = 'x';
				res[j + 2 * row][k] = 'y';
				res[j + 3 * row][k] = 'z';
			}
			k++;
			row *= 4;
			break;
		}
	}
	for (int i = 0; i < count; i++){
		res[i][k] = '\0';
	}
	*returnSize = count;
	return res;
}

int main(){
	char* digits=(char*)"23";
	int* returnSize=(int*)malloc(sizeof(int)*1);
	int length=sizeof(digits)/sizeof(char);
	char** res=letterCombinations(digits,returnSize);
	
	for(int i=0;i<*returnSize;i++){
		for(int j=0;j

2、建立字典对应（原理一样）
主要是找到正确的间隔。看了评论区的讲解，修改的一份C的代码。

char** letterCombinations(char* digits, int* returnSize) {
	if (!strcmp(digits, "") || digits == NULL) {				//空直接返回 
		*returnSize = 0;
		return NULL;
	}

	const char* str[8] = { "abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };
	int inputLength = strlen(digits);							//输入的长度 
	int* number = (int*)malloc(sizeof(int) * inputLength);
	int outputLength = 1;										//输出的个数 
	for (int i = 0; i < inputLength; i++) {						//得出最终输出个数 
		number[i] = digits[i] - '0';							//相对于 0 的差值（ASCII码） 
		outputLength *= strlen(str[number[i] - 2]);				//从 2 开始的 
	}
	char** ch = (char**)malloc(sizeof(char*) * outputLength);	//分配二级指针空间 
	for (int i = 0; i < outputLength; i++) {
		ch[i] = (char*)malloc(sizeof(char) * (inputLength + 1));//分配一级指针空间，多分配一个放截至符 
	}

	int k = 1;
	for (int i = 0; i < inputLength; i++) {
		int len = strlen(str[number[i] - 2]);					//定义当前数字代表几个字符 
		k *= len;
		int block = outputLength / k;							//计算间隔 
		for (int j = 0; j < k; j++) {							//依次加入当前数字的字符 
			for (int r = j * block; r < (j + 1) * block; r++) {	//按间隔赋值 
				ch[r][i] = str[number[i] - 2][j % len];			
			}
		}
	}

	for (int i = 0; i < outputLength; i++) {					//截至符 
		ch[i][inputLength] = '\0';
	}
	*returnSize = outputLength;
	return ch;
}

调试

#include
#include
#include

int main() {
	char* digits = (char*)"23";
	int* returnSize = (int*)malloc(sizeof(int) * 1);
	int length = sizeof(digits) / sizeof(char);
	char** res = letterCombinations(digits, returnSize);

	for (int i = 0; i < *returnSize; i++) {
		for (int j = 0; j < length; j++) {
			printf("%c", res[i][j]);
		}
		printf("\n");
	}
	return 0;
}

结果