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:
    前后端交互, 安装drf, restful接口规范, pycharm断点调试
    django中文设置, axios, CORS, 全局js配置, Vue配置jq + bs
    js原型, Vue项目环境搭建, Vue项目目录结构, Vue项目生命周期, 小组件使用, 全局样式, 路由跳转, 组件的生命周期钩子, 路由传参
    Vue组件
    Vue表单指令, 条件指令, 循环指令, 成员
    question1 赋值运算操作符
    CH15 面向对象程序设计
    CH12 动态内存
    CH11 关联容器
    CH10 泛型算法
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4582179.html
Copyright © 2011-2022 走看看