leetCode 17. Letter Combinations of a Phone Number 字符串 | 回溯 | Medium

17. 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"].

题目大意:

输入一个数字字符串。每一个数字对应手机键盘的数字键,每一个数字键对应几个字母,数字字符串的长度个字母组成一个结果串,求出这些结果串。

思路:

当看到这个题的时候,第一眼就会想要回溯算法。全部遍历。

代码如下:

class Solution {
public:
    void combine(vector<vector<char> > &all,string &digits,int index,string &temp,vector<string> &result)
    {
        if(index == digits.size())
        {
            result.push_back(temp);
            return ;
        }
        else
        {
            for(int i = 0 ; i < all[digits[index] - ‘0‘].size(); i++)//典型的回溯
            {
                temp.push_back(all[digits[index] - ‘0‘][i]);//加入新元素
                combine(all,digits,index+1,temp,result);//进入下一次递归
                temp.pop_back();//恢复到修改之前的状态
            }
        }
    }
    
    vector<string> letterCombinations(string digits) {
        vector<string> result;//存放结果的容器
        if(digits.empty())
            return result;
        vector<vector<char> > all;//字典
        string temp;//存放临时串
        all.push_back(vector<char>{});//0没有字母
        all.push_back(vector<char>{});//1没有字母
        all.push_back(vector<char>{‘a‘,‘b‘,‘c‘});//2
        all.push_back(vector<char>{‘d‘,‘e‘,‘f‘});//3
        all.push_back(vector<char>{‘g‘,‘h‘,‘i‘});//4
        all.push_back(vector<char>{‘j‘,‘k‘,‘l‘});//5
        all.push_back(vector<char>{‘m‘,‘n‘,‘o‘});//6
        all.push_back(vector<char>{‘p‘,‘q‘,‘r‘,‘s‘});//7
        all.push_back(vector<char>{‘t‘,‘u‘,‘v‘});//8
        all.push_back(vector<char>{‘w‘,‘x‘,‘y‘,‘z‘});//9
        
        combine(all,digits,0,temp,result);
        
        return result;
    }
};

回溯的思路再复习:

  1. 找到递归出口。
  2. 循环
  3. 递归进入

    (1).对初始值修改

    (2).进入下一递归

    (3).恢复对初始值的修改

2016-08-18 22:11:15

时间: 2024-10-11 01:48:22

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