腾讯2017暑期实习生编程题

#include <iostream>
using namespace std;
#include <string>
#include <algorithm>
#include <memory.h>
const int MAX_NUM = 1010;
int dp[MAX_NUM][MAX_NUM];

//先求s的反串reverse，然后求他们的最长的公共子序列，要删除的字符个数就能知道
//时间复杂度O(N^2)
int getRemoveNumber(string &str)
{
    string s(str);
    reverse(s.begin(), s.end());
    memset(dp, 0, sizeof(dp));
    int val = str.length();
    int val1 = str.size();
    for (int i = 0; i < s.size(); i++)
    {
        for (int j = 0; j < s.size(); j++)
        {
            if (str[i] == s[j])
            {
                dp[i + 1][j + 1] = dp[i][j] + 1;
            }
            else
                dp[i + 1][j + 1] = max(dp[i][j+1],dp[i+1][j]);
        }
    }
    return s.size() - dp[s.size()][s.size()]-1;  //size()和length的区别
}

int main()
{
    string str;
    while (getline(cin,str))
    {
        cout << getRemoveNumber(str) << endl;
    }
    return 0;
}



#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;

//更简洁的方法实现
int main() {
    string s;
    while (cin >> s) {
        int m = s.size();
        vector<vector<int>> dp(m, vector<int>(m, 0));
        for (int len = 1; len < m; ++len)
        for (int i = 0, j = 0; i + len < m; ++i) {
            j = i + len;
            if (s[i] == s[j])
                dp[i][j] = dp[i + 1][j - 1];
            else
                dp[i][j] = min(dp[i + 1][j], dp[i][j - 1]) + 1;
        }
        cout << dp[0][m - 1] << endl;
    }
}

//题目描述
//
//给定一个字符串，问是否能通过添加一个字母将其变为回文串。
//
//输入描述 :
//一行一个由小写字母构成的字符串，字符串长度小于等于10。
//
//
//输出描述 :
//输出答案(YES\NO).
//
//输入例子 :
//     coco
//
// 输出例子 :
//YES

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;

bool isPalindrome(string str)
{
    string temp = str;
    reverse(str.begin(), str.end());
    return str == temp;
}

int main()
{
    string str;
    while (cin >> str)
    {
        if (str.size() <= 0 || str.size() == 1)
        {
            cout << "YES" << endl;
        }
        else
        {
            bool temp = true;
            for (size_t i = 0; i < str.size(); i++)
            {
                string strTemp = str;
                strTemp.erase(strTemp.begin() + i);   //  tmp = str.substr(0, i) + str.substr(i + 1);  
                if (isPalindrome(strTemp))
                {
                    cout << "YES" << endl;
                    temp = false;
                    break;
                }
            }
            if (temp)
            {
                cout << "NO" << endl;
            }
        }
    }
    return 0;
}



////LCS实现

#include <stdio.h>
#include <string.h>
#define MAXLEN 100

void LCSLength(char *x, char *y, int m, int n, int c[][MAXLEN], int b[][MAXLEN])
{
    int i, j;
    for (i = 0; i <= m; i++)
        c[i][0] = 0;
    for (j = 1; j <= n; j++)
        c[0][j] = 0;
    for (i = 1; i <= m; i++)
    {
        for (j = 1; j <= n; j++)
        {
            if (x[i - 1] == y[j - 1])
            {
                c[i][j] = c[i - 1][j - 1] + 1;
                b[i][j] = 0;
            }
            else if (c[i - 1][j] >= c[i][j - 1])
            {
                c[i][j] = c[i - 1][j];
                b[i][j] = 1;
            }
            else
            {
                c[i][j] = c[i][j - 1];
                b[i][j] = -1;
            }
        }
    }
}

void PrintLCS(int b[][MAXLEN], char *x, int i, int j)
{
    if (i == 0 || j == 0)
        return;
    if (b[i][j] == 0)
    {
        PrintLCS(b, x, i - 1, j - 1);
        printf("%c ", x[i - 1]);
    }
    else if (b[i][j] == 1)
        PrintLCS(b, x, i - 1, j);
    else
        PrintLCS(b, x, i, j - 1);
}

int main(int argc, char **argv)
{
    char x[MAXLEN] = { "ABCBDAB" };
    char y[MAXLEN] = { "BDCABA" };
    int b[MAXLEN][MAXLEN];
    int c[MAXLEN][MAXLEN];
    int m, n;

    m = strlen(x);
    n = strlen(y);

    LCSLength(x, y, m, n, c, b);
    PrintLCS(b, x, m, n);

    return 0;
}