【LeetCode】17. 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.

 

枚举所有情况。

对于每一个输入数字,对于已有的排列中每一个字符串,分别加入该数字所代表的每一个字符。

所有是三重for循环。

举例:

初始化排列{""}

1、输入2,代表"abc"

已有排列中只有字符串"",所以得到{"a","b","c"}

2、输入3,代表"def"

(1)对于排列中的首元素"a",删除"a",并分别加入'd','e','f',得到{"b","c","ad","ae","af"}

(2)对于排列中的首元素"b",删除"b",并分别加入'd','e','f',得到{"c","ad","ae","af","bd","be","bf"}

(2)对于排列中的首元素"c",删除"c",并分别加入'd','e','f',得到{"ad","ae","af","bd","be","bf","cd","ce","cf"}

注意

(1)每次添加新字母时,应该先取出现有ret当前的size(),而不是每次都在循环中调用ret.size(),因为ret.size()是不断增长的。

(2)删除vector首元素代码为:

ret.erase(ret.begin());

 

 

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> dict(10);
        dict[0] = " ";
        dict[1] = " ";
        dict[2] = "abc";
        dict[3] = "def";
        dict[4] = "ghi";
        dict[5] = "jkl";
        dict[6] = "mno";
        dict[7] = "pqrs";
        dict[8] = "tuv";
        dict[9] = "wxyz";
        
        vector<string> ret;
        ret.push_back("");
        for(int i = 0; i < digits.size(); i ++)
        {//for each digit
            int size = ret.size();
            for(int j = 0; j < size; j ++)
            {//for existing string
                string head = ret[0];
                ret.erase(ret.begin());
                for(int k = 0; k < dict[digits[i]-'0'].size(); k ++)
                {//appending
                    ret.push_back(head + dict[digits[i]-'0'][k]);
                }
            }
        }
        return ret;
    }
};

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