中国电信IT研发中心 2019校园招聘笔试F卷 编程题-2018.09.10

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1

#include <bits/stdc++.h>
using namespace std;
int arr[26];
int main()
{
    string str;
    cin >> str;
    vector<int> vec;
    for(int i=0; i<str.size(); i++){
        int temp = str[i]-'a';
        arr[temp]++;
    }
    int minNum = 0;
    //找出最少的字符  一定要排除字符数量为0的情况
    for(int i=0; i<26; i++){
        if(arr[i] != 0){
            minNum = i;
            break;
        }
    }
    for(int i=0; i<26; i++){
        if(arr[i] < arr[minNum] && arr[i] != 0)
            minNum = i;
    }
    //将数量最少的字符对应的数字全部加入vec
    for(int i=0; i<26; i++){
        if(arr[i] == arr[minNum])
            vec.push_back(i);
    }
    for(int i=0; i<str.size(); i++){
        bool flag = true;
        int temp = str[i]-'a';
        for(int j=0; j<vec.size(); j++){//判断字符是否主要删除
            if(vec[j] == temp){
                flag = false;
                break;
            }

        }
        if(flag)
            cout << str[i];
    }
    return 0;
}

2
剑指offer原题 牛客网题解 点击链接

#include <iostream>
#include <vector>
#define min(a, b) (a<b ? a : b);
using namespace std;
int getUglyNumber(int index) {
    if(index < 7)
        return index;
    vector<int> res(index);
    res[0] = 1;
    int t2 = 0, t3 = 0, t5 = 0, i;
    for(i = 1; i < index; ++i){
        int minTemp = min(res[t2]*2, res[t3]*3);
        res[i] = min(res[t5]*5, minTemp);
        if(res[i] == res[t2]*2)
            t2++;
        if(res[i] == res[t3]*3)
            t3++;
        if(res[i] == res[t5]*5)
            t5++;
    }
    return res[index - 1];
}

int main()
{
    int n;
    cin >> n;
    cout << getUglyNumber(n) << endl;
    return 0;
}

3

KMP扩展
参考链接1
参考链接2

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转载自blog.csdn.net/whl_program/article/details/82589620