[DFS]电话号码的字母组合

电话号码的字母组合

这道题目的本质是笛卡尔积问题，可以通过DFS求解，分别采用C/Java语言实现

#include <stdio.h>
#include <stdlib.h>

char *phoneMap[10] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
static char **combinations; // 记录最终结果
static int combinations_size;

static char *combination; // 记录每次回溯的字符串
static int combination_size;

void backtrack(char *digits, int size, int index) {
    if (index == size) {
        char *tmp = (char *) malloc(sizeof(char) * (combination_size + 1));
        memcpy(tmp, combination, sizeof(char) * (combination_size + 1));
        combinations[combinations_size++] = tmp;
        return;
    }
    char *letters = phoneMap[digits[index] - '0'];
    int len = strlen(letters);
    for (int i = 0; i < len; i++) {
        combination[combination_size++] = letters[i];
        combination[combination_size] = '\0'; // 记录结束标识
        backtrack(digits, size, index + 1);
        combination[--combination_size] = '\0'; // 状态复原
    }
}

char **letterCombinations(char *digits, int *returnSize) {
    combinations_size = combination_size = 0; // 在入口函数初始化0，不然leetcode编译报错
    int digits_size = strlen(digits);
    if (digits_size == 0) {
        *returnSize = 0;
        return combinations;
    }
    int num = 1;
    for (int i = 0; i < digits_size; i++) num *= 4;
    combinations = (char **) malloc(sizeof(char *) * num);
    combination = (char *) malloc(sizeof(char *) * digits_size);
    backtrack(digits, digits_size, 0);
    *returnSize = combinations_size;
    return combinations;
}

int main() {
    char *str = "23";
    int ret = 0;
    char **res = letterCombinations(str, &ret);
    for (int i = 0; i < ret; i++) {
        printf("%s\n", *(res + i));
    }
}