leetcode 17. Letter Combinations of a Phone Number-优快云博客

本文链接：https://blog.youkuaiyun.com/hello_my_coder/article/details/78081321

本文介绍了一种通过递归算法生成电话号码所有可能的字母组合的方法。针对输入的数字字符串，利用递归函数逐层遍历对应的字母，最终输出所有可能的字母组合。

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题目描述：
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.

讲道理一开始还没看懂，就是按照手机的数字键盘上面对应的字母，将数字串转换为可能的字母串。看了下面这个表就明白了。

1       2(abc)  3(def)
4(ghi)  5(jkl)  6(mno)
7(pqrs) 8(tuv)  9(wxyz)
*       0       #

应该不难想到可以用递归的方法一层一层循环的遍历。时间复杂度是O(n*m)，以下是解答的方法：

class Solution {
private:
    vector<string> result;
public:
    vector<string> letterCombinations(string digits) {
        if (digits == "") {
            return result;
        }
        recursion(digits, 0, "");
        return result;
    }
    void recursion(string digits, int i,string now) { 

        string digitToletter[8] = { "abc", "def","ghi","jkl","mno","pqrs","tuv","wxyz"};

        int letterNum = 3;
        if (digits[i] == '7' || digits[i] == '9') {
            letterNum = 4;
        }
            for (int k = 0; k < letterNum; k++) {
                if (i == 0) {
                    now = "";
                }
                now = now + digitToletter[digits[i]-'2'][k];
                if (i == digits.length() - 1) {
                    result.push_back(now);
                }
                else {
                    recursion(digits, i + 1, now);
                }
                //这里的两次字符串长度减一是十分关键的，
                //这里是遍历完一个数字对应的一个字母后，字符串长度减一以便继续遍历
                now = now.substr(0, now.length() - 1);
            }
            //这里是遍历完一个数字对应的所有字母后，回到上一层的数字进行遍历
            now = now.substr(0, now.length() - 1);
        }

};