力扣 题目17-- 电话号码的字母组合

题目

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题解

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这题我们可以发现用四层循环(即循环第一个数字的字母然后里面套着第二个数字的字母 全部遍历然后输出)应该可以解决这个问题

但是既然是算法题还是中等我们应该另辟蹊径(好像可以用回溯算法 但是这里我这里采用了其他暴力?方法)

观察结果

输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]

那么我们可不可以先把第一个数字的字母全部放进去 然后再去循环第二个呢

即 第一轮循环 

输出：["a","a","a","b","b","b","c","c","c"]

第二轮循环 

输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]

1.先创建一个vector

vector<string> Telephone = { "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };

这样下标就是对应数字

2.先获取一下最后结果有多少种可能性即 数字代表字母个数相乘 由于别忘了用resize将返回的result扩大到number 个空间

        int number = 1;
        for (int i = 0; i < digits.size(); i++)
        {
            number = number * Telephone[digits[i] - 48].size();
        }
        result.resize(number);
        int number2 = number;

3.遍历

先固定digits的数字->再判断要循环的次数(看看是否到number 不是就继续循环)->然后循环一遍数字代表的字符串->将字符串放进对应的空间

for (int i = 0; i < digits.size(); i++) {
            int size = Telephone[digits[i] - 48].size();
            for (int h = 0; h < number2;) {
                for (int j = 0; j < size; j++) {
                    for (int g = (number / size) * j+h; g < (number / size) * (j + 1) + h; g++) {
                        result[g] = result[g] + Telephone[digits[i] - 48][j];
                    }
                }
                h = h + number;
            }
            number = number / Telephone[digits[i] - 48].size();
        }

4.如果为空需要直接返回

vector<string> result = {};
if (digits == "") {
            return result;
        }

代码

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#include<iostream>
#include<string>
#include<vector>
using namespace std;
vector<string> Telephone = { "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };
class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> result = {};
        if (digits == "") {
            return result;
        }
        int number = 1;
        for (int i = 0; i < digits.size(); i++)
        {
            number = number * Telephone[digits[i] - 48].size();
        }
        result.resize(number);
        int number2 = number;
        for (int i = 0; i < digits.size(); i++) {
            int size = Telephone[digits[i] - 48].size();
            for (int h = 0; h < number2;) {
                for (int j = 0; j < size; j++) {
                    for (int g = (number / size) * j+h; g < (number / size) * (j + 1) + h; g++) {
                        result[g] = result[g] + Telephone[digits[i] - 48][j];
                    }
                }
                h = h + number;
            }
            number = number / Telephone[digits[i] - 48].size();
        }
        return result;
    }
};
int main() {
    Solution sol;
    string digits = "789";
    vector<string> result = sol.letterCombinations(digits);
    for (vector<string> ::iterator it = result.begin(); it != result.end(); it++) {
        cout << *it << endl;
    }
}