电话号码对应英语单词

最新推荐文章于 2019-08-29 21:13:20 发布

原创最新推荐文章于 2019-08-29 21:13:20 发布 · 1.2k 阅读

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#电话 #numbers #keyboard #null

博客探讨了如何将电话号码5869872转换为对应的字母组合，通过对比编程之美中复杂的多重循环解法与书中的递归程序，强调了递归解法的简洁高效，无需额外保存信息。

对于号码5869872，可以依次输出其代表的所有字母组合，如JTMWTPA、JTMWTPB.....

#include <iostream>
#include <string>
#include <vector>

using namespace std;

char keyBoard[10][10] =
{
	"",
	"",
	"abc",
	"def",
	"ghi",
	"jkl",
	"mno",
	"pqrs",
	"tuv",
	"wxyz"
};

int keyLength[10] = {0, 0, 3, 3, 3, 3, 3, 4, 3, 4};
vector<char> result;

void printCombine(const char *numbers, int index)
{
	if (numbers == NULL)
		return;

	if (index == strlen(numbers))
	{
		copy(result.begin(), result.end(), ostream_iterator<char>(cout, " "));
		cout << endl;
		return;
	}


	for (int i = 0; i < keyLength[numbers[index] - '0']; i++)
	{
		result.push_back(keyBoard[numbers[index] - '0'][i]);
		printCombine(numbers, index + 1);
		result.pop_back();
	}
}

void main()
{
	printCombine("422", 0);
}

编程之美上的解法，多重for循环，看得头晕。。。

void printCombine(const char *numbers)
{
	int answer[2];
	for (answer[0]= 0; answer[0] < keyLength[numbers[0] - '0']; answer[0]++)
	{
		for (answer[1] = 0; answer[1] < keyLength[numbers[1] - '0']; answer[1]++)
		{
			for (int i = 0; i < 2; i++)
				cout << keyBoard[numbers[i] - '0'][answer[i]];
			cout << endl;
		}
	}
}

根据递归解法，可以想象本题的实质不过是进制的变种而已，前面的习题是10进制，这个是N进制，不同位的最大值不一样，判断并取进位

void printCombine2(int *numbers, int size)
{
	if (numbers == NULL || size <= 0)
		return;

	int *index = new int[size];
	if (index == NULL)
		return;

	for (int i = 0; i < size; i++)
		index[i] = 0;

	int pos = size - 1;
	bool flag = false;
	while (true)
	{
		if (pos < 0)
			break;

		for (int i = 0; i < size; i++)
			printf("%c", keyBoard[numbers[i]][index[i]]);
		printf("\n");

		index[size - 1]++;
		for (pos = size - 1; pos >= 0; pos--)
		{
			if (flag == true)
				index[pos]++;

			if (index[pos] == keyLength[numbers[pos]])
			{
				index[pos] = 0;
				flag = true;
				continue;
			}
			else
			{
				flag = false;
				break;
			}
		}
	}
}

不过编程之美上给出了似乎更为简洁的解答，其实实质一样

void printCombine2(int *numbers, int size)
{
	if (numbers == NULL || size <= 0)
		return;

	int *index = new int[size];
	if (index == NULL)
		return;

	for (int i = 0; i < size; i++)
		index[i] = 0;

	int pos = size - 1;
	bool flag = false;
	while (true)
	{
		if (pos < 0)
			break;

		for (int i = 0; i < size; i++)
			printf("%c", keyBoard[numbers[i]][index[i]]);
		printf("\n");

		index[size - 1]++;
		for (pos = size - 1; pos >= 0; pos--)
		{
			if (flag == true)
				index[pos]++;

			if (index[pos] == keyLength[numbers[pos]])
			{
				index[pos] = 0;
				flag = true;
				continue;
			}
			else
			{
				flag = false;
				break;
			}
		}
	}
}

还是书上给出的递归程序靠谱，不需要保存额外信息：

void printCombine3(int *numbers, int *answer, int index, int size)
{
	if (numbers == NULL || size <= 0)
		return;

	if (index == size)
	{
		for (int i = 0; i < size; i++)
		{
			printf("%c", keyBoard[numbers[i]][answer[i]]);
		}
		printf("\n");
		system("pause");
		return;
	}

	for (answer[index] = 0; answer[index] < keyLength[numbers[index]]; answer[index]++)
	{
		printCombine3(numbers, answer, index + 1, size);
	}
}