[C++]LeetCode: 87 Letter Combinations of a Phone Number

标签：backtracking   string   leetcode   

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.

Answer 1: 递归方法(DFS)

思路：我们按照数字字符串的位数，逐一DFS拆解。得到第一位对应的字母之后，再继续DFS搜索，直到达到数字的位数为止。添加这个结果，并返回。

Attention：

1. 如何初始化数字字母查找表，用字符串数组表示。字符串数组一维坐标代表定位到第几个字符串，二维坐标代表定位到字符。如numap[3][2] = ‘f‘

string numap[] = {" ","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};

 string tmpstr(digits.size(), ' ');

 tmpstr[index] = numap[digits[index] - '0'][i];

AC Code:

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> ret;
      //  if(digits.size() == 0) return ret;
        string tmpstr(digits.size(), ' ');
        letterCombinations_helper(digits, 0, tmpstr, ret);
        return ret;
    }
    
private:
    vector<vector<char> > phonetable;
    void letterCombinations_helper(string digits, int index, string tmpstr, vector<string>& ret)
    {
        string numap[] = {" ","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        if(index == digits.size())
        {
            ret.push_back(tmpstr);
            return;
        }

        for(int i = 0; i < numap[digits[index] - '0'].size(); i++)
        {
            tmpstr[index] = numap[digits[index] - '0'][i];
            letterCombinations_helper(digits, index+1, tmpstr, ret);
           // tmpstr.pop_back();
        }
        
        return;
    }
};

思路：我们按照digits的位数，迭代的去求解结果，先从第一位的所有可能开始，不断扩充结果。思路有点类似于求subset.

举个例子，比如我们要求“23”，初始化ret = [""], 2对应三个字母，所以tmp会push三次，得到tmp[0] = "" + ‘a’; tmp[1] = "" + ‘b‘; tmp[2] = "" + ‘c‘. ret = tmp; (第一次最外层迭代结束)； ret.size() = 3. tmp[0] = "a" + ‘d‘; tmp[1] = "a"+‘e‘; tmp[2] = "a"+‘f‘; tmp[3] = "b"+‘d’；.......直到得到所有的结果。（外层迭代结束）。每次都要更新目前为止的ret的所有结果, 下次迭代再在此基础上添加更多可能。

这种方法可能理解上有点难度，不太符合正常的思考过程，不过可以拓展编程思路。

Attention: 三层循环，最外层：digits的位数（digits.size()）; 第二层: 当前ret内的子集数目（ret.size()）; 最内层: 这个数字对应的所有可能字母（numap[digits[i] - ‘0‘].size()). 是BFS思路。

AC Code:

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> ret(1, "");
        string numap[] = {" "," ","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        
        for(int i = 0; i < digits.size(); i++)
        {
            vector<string> tmp;
            
            for(int j = 0; j < ret.size(); j++)
            {
                for(int k = 0; k < numap[digits[i]-'0'].size(); k++)
                {
                    tmp.push_back(ret[j] + numap[digits[i]-'0'][k]);
                }
            }
            //更新ret
            ret = tmp;
        }
        
        return ret;
    }
};

[C++]LeetCode: 87 Letter Combinations of a Phone Number

标签：backtracking   string   leetcode   

原文地址：http://blog.csdn.net/cinderella_niu/article/details/42610531