[LeetCode][JavaScript]Letter Combinations of a Phone Number

Letter Combinations of a Phone Number 

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

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

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

https://leetcode.com/problems/letter-combinations-of-a-phone-number/

 

 

 

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提示说是高贵的回溯，说白了就是递归...

思路很简单，比如给定4个数字，就是个4重循环。

但是几重循环是动态的，递归嵌套一下就好了。

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

    var target = [];
    var i, j;
    for(i in digits){
        target.push(dict[digits[i]]);
    }
    return combinations(target);

    function combinations(arr){
        var newArr = arr.slice(0); //clone
        if(newArr.length === 0){
            return [];
        } else {
            var res = [];
            var top = newArr.shift();
            var ret = combinations(newArr);
            if(ret.length === 0){
                res = top;
            }else{
                for(i = 0; i < top.length; i++){
                    for(j = 0; j < ret.length; j++){
                        res.push(top[i] + ret[j]);
                    }
                }
            }
            return res;
        }
    }
};