面试过的问题

京东编程：找幸运数

#include <iostream>
using namespace std;

int f(int x)
{
    int sum = 0;
    while (x != 0)
    {
        sum += x % 10;
        x /= 10;
    }
    return sum;
}

int g(int x)
{
    int sum = 0;
    int flag = 1;
    while (flag)
    {
        if (flag&x)
            sum++;
        flag = flag << 1;
    }
    return sum;
}

int main()
{
    int n;
    cin >> n;
    int *p = new int[n];
    for (int i = 0; i < n; i++)
    {
        cin >> p[i];
    }
    for (int i = 0; i < n; i++)
    {
        int count = 0;
        for (int j = 1; j <= p[i]; j++)
        {
            if (f(j) == g(j))
                count++;
        }
        cout << count << endl;
    }
    system("pause");
    return 0;
}

无序数组中找出和为N的两个数 Two Sum

Hash方法：可以考虑使用hash。将数组的每个元素以元素值位key，下标为value存入hash表。然后第二次遍历时，在hash中查询target做减法后的差值，hash表中存在该差值就算找到了。注意一点：当数组中存在重复元素时，注意hash表该如何查找；两个数不能为同一个数。

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        unordered_map<int, int> hashtable;
        vector<int> result;
        int n = numbers.size();
        for(int i=0;i<n;i++)
            hashtable[numbers[i]] = i+1;
            
        for(int i=0;i<n;i++)
        {
            int other = target - numbers[i];
            if(hashtable.find(other) != hashtable.end())
            {
                int j = hashtable[other];
                if(i+1 == j)
                    continue;
                result.push_back(i+1);
                result.push_back(j);
                break;
            }
        }
        return result;
    }
};

滴滴

二叉搜索树合法性，leetcode：http://www.nowcoder.com/practice/fd7f880072914464a13b89af242c0ce5?tpId=46&tqId=29080&rp=3&ru=/ta/leetcode&qru=/ta/leetcode/question-ranking

中序，判断是否递增，不要用递归，容易出错

class Solution {
public:
    void inOrderTree(TreeNode *root,vector<int> &vec)
    {
        if(root == NULL)
            return;
        inOrderTree(root->left,vec);
        vec.push_back(root->val);
        inOrderTree(root->right,vec);
    }
    bool isValidBST(TreeNode *root) {
        if(root == NULL)
            return true;
        vector<int> vec;
        inOrderTree(root,vec);
        int len = vec.size();
        if(len == 1)
            return true;
        for(int i = 0; i < len - 1;i++)
        {
            if(vec[i] >= vec[i+1])
                return false;
        }
        return true;
    }
};

 滴滴

合并两个无序数组到有序，并去重

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

int partition(int *array, int low, int high)
{
    int pivot = array[low];
    while (low < high)
    {
        while (low < high && array[high] >= pivot)
            high--;
        swap(array[low], array[high]);
        while (low < high && array[low] <= pivot)
            low++;
        swap(array[low], array[high]);
    }
    return low;
}


void QuickSort(int *nums, int low, int high)
{
    int index;
    if (low < high)
    {
        index = partition(nums, low, high);
        QuickSort(nums, low, index - 1);
        QuickSort(nums, index + 1, high);
    }
}

vector<int> mergeArray(int *a, int len_a, int *b, int len_b)
{
    QuickSort(a, 0, len_a - 1);
    QuickSort(b, 0, len_b - 1);
    vector<int> result;
    int i = 0;
    int j = 0;
    while(i < len_a && j < len_b)
    {
        if (a[i] < a[j])
        {
            result.push_back(a[i]);
            i++;
        }
        else if (a[i] > a[j])
        {
            result.push_back(a[j]);
            j++;
        }
        else
        {
            int tmp = a[i];    //处理重复元素
            result.push_back(tmp);
            i++; j++;
            while (i < len_a && a[i] == tmp)
                i++;
            while (j < len_b && b[j] == tmp)
                j++;
        }
    }
    while (i < len_a)
    {
        result.push_back(a[i]);
        i++;
    }
    while (j < len_b)
    {
        result.push_back(b[j]);
        j++;
    }
    return result;
}

int main()
{
    int a[] = { 3,2,1};
    int b[] = { 3, 33, 2, 1 };
    
    vector<int> vec = mergeArray(a, 3, b, 4);

    system("pause");
    return 0;
}

中兴，用两个线程，从1+100

#include <iostream>
#include <vector>
using namespace std;
#include <thread>

void fun1(int &sum)
{
    for (int i = 1; i <= 50; i++)
        sum += i;
}

void fun2(int &sum)
{
    for (int i = 51; i <= 100; i++)
        sum += i;
}

int main() 
{
    int sum1 = 0;
    int sum2 = 0;
    thread t1(fun1, ref(sum1));
    thread t2(fun2, ref(sum2));
    t1.join();
    t2.join();
    cout << sum1 + sum2 << endl;

    system("pause");
    return 0;
}