mysql常用查询：group by，左连接，子查询，having where

时间：2014-04-27 21:17:59 收藏：0 阅读：614

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.

从组合的角度来看，无非就是把每个位置的数字所能表示的字母都遍历一次，从而得到一个全组合的集合。这类问题可以用backtrace的方法来解决，下边的解法1和解法2分别用了非递归和递归两种方式来实现backtrace。

第二个递归的很直观，没啥好说的；第一个非递归的可以看作是从最右边的一个数字开始每次加一递增（取下一个可能的字幕就是加1），取完当前数字对应的字母就认为是该进位了 -- （下次该开始处理它左边的一位数字并同时把当前数字所取得对应字母归位 -- 重新从第一个字母开始），整个过程可以看作是这样一个数字的递增遍历过程。

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> ret;
        int len = digits.length();
        
        string n2s[] = {
            "",
            "",
            "abc",
            "def",
            "ghi",
            "jkl",
            "mno",
            "pqrs",
            "tuv",
            "wxyz"
        };
        
        // 1
        /*
        vector<int> curpos(len, 0);
        while (true) {
            ret.push_back("");
            int ind = ret.size() - 1;
            for (int i = 0; i < len; i++) {
                int nind = digits[i] - ‘0‘;
                if (n2s[nind].length() > 0) {
                    ret[ind].push_back(n2s[nind][curpos[i]]);
                }
            }
            
            int k = len - 1;
            while (k >= 0) {
                int nind = digits[k] - ‘0‘;
                if (curpos[k] < n2s[nind].length() - 1) {
                    curpos[k]++;
                    break;
                }
                
                curpos[k--] = 0;
            }
            
            if (k < 0) break;
        }
        */
        
        // 2
        string curs = "";
        genLetterCombination(ret, curs, digits, n2s, 0);
        
        return ret;
    }

private:
    void genLetterCombination(vector<string> &ret, string &curs, string &d, string n2s[], int cpos) {
        if (cpos == d.length()) {
            ret.push_back(curs);
            return;
        }
        
        int dind = d[cpos] - ‘0‘;
        for (int i = 0; i < n2s[dind].length(); i++) {
            curs.push_back(n2s[dind][i]);
            genLetterCombination(ret, curs, d, n2s, cpos + 1);
            curs.pop_back();
        }
    }
};