Letter Combinations of a Phone Number
来源:互联网 发布:车牌编码软件 编辑:程序博客网 时间:2024/06/17 08:08
每日算法——leetcode系列
问题 3Sum Closest
Difficulty: Medium
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.
class Solution {public: vector<string> letterCombinations(string digits) { }};
翻译
电话号码的字母组合
难度系数:中等
给定一个数字组成的字符串, 返回这个数字所有所有可能代表的字母组合。
每个数字代表的字母(像电话按键)如下面图显示一样
输入:数字字符串 "23"输出: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
注意:
虽然上面的的答案是按字典顺序的,你的答案可以是你想要的任意顺序。
思路
这题很像 《编程珠玑》 2.6习题及 《编程之美》 3.2节讲的东西
这题不用找出单词, 只是返回字母组合, 相对要简单。
- map表
这题还有一个map字眼,可以把按键对应的数字组成如下的一个二维数组的map表phoneNum。
0 {' ', '\0', '\0', '\0'} 1 {'\0', '\0', '\0', '\0'} 2 {'a', 'b', 'c', '\0'} 3 {'d', 'e', 'f', '\0'} 4 {'g', 'h', 'i', '\0'} 5 {'j', 'k', 'l', '\0'} 6 {'m', 'n', 'o', '\0'} 7 {'p', 'q', 'r', 's'} 8 {'t', 'u', 'v', '\0'} 9 {'w', 'x', 'y', 'z'}
- 输入的字符串根据map表取出按键上的字符
遍历每个输入的数字字符串中的每个数字(当然每次遍历的时候要判断这个是否为数字isdigit),根据map表取出相应按键上的字符依次插入到result中 - 组合
再循环执行第二步,但要组合。
例如
输入为“2”的时候, result = {“a”, “b”, “c”}
输入为“23”的时候:
第二步中第一次 result = {“a”, “b”, “c”}
第二次 phoneNum[3]= {“d”, “e”, “f”}, result[i] + phoneNum[3][i]就好
代码
class Solution {public: vector<string> letterCombinations(string digits) { vector<string> result; if (digits.empty()) { return result; } // 电话号码字母map表 char phoneNum[10][4] = { {' ', '\0', '\0', '\0'}, {'\0', '\0', '\0', '\0'}, {'a', 'b', 'c', '\0'}, {'d', 'e', 'f', '\0'}, {'g', 'h', 'i', '\0'}, {'j', 'k', 'l', '\0'}, {'m', 'n', 'o', '\0'}, {'p', 'q', 'r', 's'}, {'t', 'u', 'v', '\0'}, {'w', 'x', 'y', 'z'}, }; for (size_t i = 0; i < digits.size(); ++i) { if (!isdigit(digits[i])) { vector<string> temp; return temp; } int digit = digits[i] - '0'; if (result.empty()) { for (int j = 0; j < 4; ++j) { if (phoneNum[digit][j] == '\0'){ break; } // 数字键对应的字符 string s; s = phoneNum[digit][j]; result.push_back(s); } continue; } vector<string> tmpResult; for (int j = 0; j < result.size(); ++j) { for (int k = 0; k < 4; ++k) { if (phoneNum[digit][k] == '\0'){ break; } string s = result[j] + phoneNum[digit][k]; tmpResult.push_back(s); } } result = tmpResult; } return result; }};
0 0
- LeetCode: Letter Combinations of a Phone Number
- LeetCode Letter Combinations of a Phone Number
- LeetCode: Letter Combinations of a Phone Number
- [Leetcode] Letter Combinations of a Phone Number
- [LeetCode] Letter Combinations of a Phone Number
- LeetCode18:Letter Combinations of a Phone Number
- Letter Combinations of a Phone Number
- 【leetcode】Letter Combinations of a Phone Number
- LeetCode: Letter Combinations of a Phone Number
- Letter Combinations of a Phone Number
- Letter Combinations of a Phone Number
- [LeetCode]Letter Combinations of a Phone Number
- LeetCode-Letter Combinations of a Phone Number
- Letter Combinations of a Phone Number
- [LeetCode] Letter Combinations of a Phone Number
- Letter Combinations of a Phone Number
- LeetCode - Letter Combinations of a Phone Number
- LeetCode:Letter Combinations of a Phone Number
- 获取MAC系统最高权限(解决文件无法拷贝问题)
- Linux Netlink通信机制详解
- 一篇实用的Latex的入门教程
- 杭电-5569-动太规划
- scikit-learn的主要模块和基本使用
- Letter Combinations of a Phone Number
- QT 模拟鼠标点击的例子
- 在MVC使用Log4Net
- linux 查看elf相关命令
- Oracle 设定允许访问的IP地址
- 简单c语言习题:矩阵求乘积
- 无线路灯项目——SIM900A调试
- 山东省第四届蓝桥杯 ///题目标题: 高斯日记//c/c++组
- opencv CvArr、Mat、CvMat、IplImage、BYTE之间的转换