17. Letter Combinations of a Phone Number

[Tcdian]

17. Letter Combinations of a Phone Number

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。

Example

Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note

• 尽管上面的答案是按字典序排列的，但是你可以任意选择答案输出的顺序。

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {string} digits
 * @return {string[]}
 */
var letterCombinations = function(digits) {
    if (digits.length === 0) {
        return [];
    }
    const dictionary = new Map([
        ['2', ['a', 'b', 'c']],
        ['3', ['d', 'e', 'f']],
        ['4', ['g', 'h', 'i']],
        ['5', ['j', 'k', 'l']],
        ['6', ['m', 'n', 'o']],
        ['7', ['p', 'q', 'r', 's']],
        ['8', ['t', 'u', 'v']],
        ['9', ['w', 'x', 'y', 'z']],
    ]);
    const result = [];
    const combination = [];
    backtracking(0);
    return result;

    function backtracking(index) {
        if (index === digits.length) {
            result.push(combination.join(''));
            return;
        }
        let letters = dictionary.get(digits[index]);
        for (let i = 0; i < letters.length; i++) {
            combination.push(letters[i]);
            backtracking(index + 1);
            combination.pop();
        }
    }
};

• TypeScript Solution

function letterCombinations(digits: string): string[] {
    if (digits.length === 0) {
        return [];
    }
    const dictionary = new Map([
        ['2', ['a', 'b', 'c']],
        ['3', ['d', 'e', 'f']],
        ['4', ['g', 'h', 'i']],
        ['5', ['j', 'k', 'l']],
        ['6', ['m', 'n', 'o']],
        ['7', ['p', 'q', 'r', 's']],
        ['8', ['t', 'u', 'v']],
        ['9', ['w', 'x', 'y', 'z']],
    ]);
    const result: string[] = [];
    const combination: string[] = [];
    backtracking(0);
    return result;

    function backtracking(index: number) {
        if (index === digits.length) {
            result.push(combination.join(''));
            return;
        }
        let letters = dictionary.get(digits[index])!;
        for (let i = 0; i < letters.length; i++) {
            combination.push(letters[i]);
            backtracking(index + 1);
            combination.pop();
        }
    }
};