zoukankan      html  css  js  c++  java
  • [LeetCode] Letter Combinations of a Phone Number

    Well, a typical backtracking problem. Make sure you are clear with the following three problems:

    1. What is a partial solution and when is it finished? --- In this problem, the partial solution is a combination and it is finished once it is of the same length as digits.
    2. How to find all the partial solutions? --- In the following code, I use a nested for-loop to traverse all the possible starting elements for each digit.
    3. When to make recursive calls? --- In the following code, once an element is added to the partial solution, we call the function to generate the remaining elements recursively.

    Of course, remember to recover to the previous status once a partial solution is done. In the following code, line 18 (comb.resize(comb.length() - 1)) is for this purpose.

    The following should be self-explanatory :)

     1 class Solution {
     2 public:
     3     vector<string> letterCombinations(string digits) {
     4         string mp[] = {" ", " ", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
     5         vector<string> res;
     6         if (digits.empty()) return res;
     7         string comb;
     8         combinations(digits, 0, mp, comb, res);
     9         return res;
    10     }
    11 private:
    12     void combinations(string& digits, int start, string mp[], string& comb, vector<string>& res) {
    13         if (comb.length() == digits.length()) {
    14             res.push_back(comb);
    15             return;
    16         }
    17         for (int i = 0; i < (int)mp[digits[start] - '0'].length(); i++) {
    18             comb += (mp[digits[start] - '0'][i]);
    19             combinations(digits, start + 1, mp, comb, res);
    20             comb.resize(comb.length() - 1);
    21         }
    22     }
    23 };

     Well, the above recursive backtracking solution is a typical solution. In fact, this problem can also be solved iteratively. Refer to this link for more information. I have rewritten the code below for your reference.

    class Solution {
    public:
        vector<string> letterCombinations(string digits) {
            vector<string> res;
            if (digits.empty()) return res;
            string mp[] = {" ", " ", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
            res.push_back("");
            for (int i = 0; i < (int)digits.length(); i++) {
                string letters = mp[digits[i] - '0'];
                vector<string> temp;
                for (int j = 0; j < (int)letters.length(); j++)
                    for (int k = 0; k < (int)res.size(); k++)
                        temp.push_back(res[k] + letters[j]);
                swap(res, temp);
            }
            return res;
        }
    };
  • 相关阅读:
    魔法方法中的__str__和__repr__区别
    新建分类目录后,点击显示错误页面?
    3.用while和for循环分别计算100以内奇数和偶数的和,并输出。
    2.for循环实现打印1到10
    1.while循环实现打印1到10
    021_for语句
    014_运算符_字符串连接
    020_while语句
    019_增强switch语句
    018_switch语句
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4582179.html
Copyright © 2011-2022 走看看