Letter Combinations of a Phone Number

[barretlee]

本题难度：★★

给定一个纯数字的字符串，根据下面九宫格表返回所有可能的字母组合：

比如给定 "23"，那么所有可能的输出为：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]

注意：上面给出的是按照辞典顺序输出的，可以不考虑顺序。

[daghlny]

const vector<string> letters = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

int dfs(string& s, int *numbers, int cur, int n)
{
    if ( n == 0 ) return 0;
    if ( cur == n ){
        cout << s << endl;
        return 0;
    }
    const string &cands = letters[numbers[cur]];
    for(int i = 0; i < cands.size(); ++i)
    {
        s.push_back(cands[i]);
        dfs(s, numbers, cur+1, n);
        s.erase(s.end()-1);
    }
    return 0;
}

int main(void)
{
    char numbers_str[100];
    scanf("%s", numbers_str);
    int *numbers = new int[strlen(numbers_str)];
    for(int i = 0; i < strlen(numbers_str); ++i)
        numbers[i] = static_cast<int>(numbers_str[i] - '0');
    string s;
    dfs( s, numbers, 0, strlen(numbers_str));
    delete[] numbers;
    return 0;
}

[dahuanggithub]

function f(a){
  var lib = ['','',"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"];
  var res = [];
  var arr = [];
  for(var i=0;i<a.length;i++){
    if(lib[a[i]]!==''){
      arr.push(lib[a[i]]);
    }
  }
  if(arr.length<2){
    if(arr.length===0){return res;}
    res = arr[0].split("");
  }else{
    res = arr.reduce(function(x,y){
     var _arr = [];
     for(var i=0;i<x.length;i++){
      for(var j=0;j<y.length;j++){
        _arr.push(x[i]+y[j]);
      }
     }
    return _arr;
    });
  }
  return res;
}
console.log(f("102"));

[2609974827]

var lujin = function(indexStr) {
	if (!(/^[0-9]+$/.test(indexStr))) {
		return '';
	}
	var dictionary = ['', "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"];
	var indexArr = indexStr.split('');
	var targetDictionary = [];
	for (var i = 0, l = indexArr.length; i < l; i++) {
		var tempArr = dictionary[indexArr[i]].split('');
		for (var j = 0, m = tempArr.length; j < m; j++) {
			tempArr[j] = {
				value: tempArr[j],
				x_axis: i,
				y_axis: j
			}
		}
		targetDictionary.push(tempArr);
	}
	var results = traversal(targetDictionary);
	console.log(targetDictionary);
	console.log(results);
};
var traversal = function(arr) {
	var results = [];
	var stockArr = new arrayStack();
	var lastTop = {}, //存储出栈元素
		temp = {}; //存储栈顶元素
	var i = 0; //避免死循环
	stockArr.push(arr[0][0]);
	while (stockArr.length() > 0) {
		i++;
		temp = stockArr.top();
		//如果到达终点记录本次路劲值
		if (temp.x_axis == arr.length - 1) {
			results.push(stockArr.toString());
		}
		//判断是否有下一层，并且下一层不是上次出栈元素
		if (arr[temp.x_axis + 1] && arr[temp.x_axis + 1][0] && lastTop.x_axis != temp.x_axis + 1) {
			stockArr.push(arr[temp.x_axis + 1][0]);
		} else {
			if (arr[temp.x_axis][temp.y_axis + 1]) {
				lastTop = stockArr.pop();
				stockArr.push(arr[temp.x_axis][temp.y_axis + 1]);
			} else {
				lastTop = stockArr.pop();
			}
		}
		//到达限制死循环点
		if (i == 1000) {
			console.log('到达限制死循环点');
			break;
		}
	}
	return results;
};
var arrayStack = function() {
	this.stock = [];
	this.push = function(ele) {
		this.stock.push(ele);
	};
	this.pop = function() {
		return this.stock.pop();
	};
	this.top = function() {
		return this.stock[this.stock.length - 1];
	};
	this.size = function() {
		return this.stock.length;
	};
	this.clear = function() {
		this.stock.splice(0);
		return true;
	};
	this.length = function() {
		return this.stock.length;
	};
	this.toString = function() {
		var tempStr = '';
		for (var i = 0, l = this.stock.length; i < l; i++) {
			tempStr += this.stock[i].value;
		}
		return tempStr;
	}
}

[barretlee]

这道题使用深度优先遍历会比较简单。停了一周多，本周继续~