2.15 给出一个有效的算法来确定在整数 A 1 < A 2 < A 3 < ⋯ < A N A_1< A_2<A_3<\dots < A_N A1​<A2​<A3​<⋯<AN​的数组是否存在整数 i i i​使得 A i = i A_i = i Ai​=i。

#include <iostream>
#include <vector>

using namespace std;

bool indexIsValue(const vector<int>& A, int low, int high)
{
    if (low == high)
    {
        if (A[low] == low)
            return true;
        return false;
    }

    int mid = (low + high) / 2;

    if (A[mid] > mid)
        return indexIsValue(A, low, mid - 1);
    else if (A[mid] < mid)
        return indexIsValue(A, mid + 1, high);
    else
        return true;
}

int main()
{
    vector<int> vec{ -1,1,3,4,6 };
    cout << indexIsValue(vec, 0, vec.size() - 1) << endl;

    system("pause");
    return 0;
}

2.16 基于下列个结论编写另外的gcd算法……

#include <iostream>
#include <vector>

using namespace std;

int gcd(int a, int b)
{
    int high = a, low = b;
    if (high % low == 0)
        return low;

    if (!high % 2 && !low % 2)
        return 2 * gcd(high / 2, low / 2);
    else if (high % 2 == 0 && low % 2 != 0)
    {
        if (high % 2 > low)
            return gcd(high % 2, low);
        else
            return gcd(low, high / 2);
    }
    else if (high % 2 != 0 && low % 2 == 0)
        return gcd(high, low / 2);
    else
        return gcd((high + low) / 2, (high - low) / 2);
}

int main()
{
    cout << gcd(4, 3) << endl;

    system("pause");
    return 0;
}

2.17 给出有效的算法

// a 最小子序列和
int minSumRec(const vector<int>& a)
{
    int minSum = a[0], thisSum = 0;

    for (size_t i = 0; i < a.size(); ++i)
    {
        thisSum += a[i];

        minSum = min(minSum, thisSum);
        thisSum = min(thisSum, 0);
    }

    return minSum;
}

// b 最小正子序列和， 目前只会这种暴力求解的方式，插个眼
int minPosSumRec(const vector<int>& a)
{
    int minPosSum = 0;

    for (int i = 0; i < a.size(); ++i)
    {
        int thisSum = 0;
        for (int j = i; j < a.size(); ++j)
        {
            thisSum += a[j];
            
            if ((minPosSum == 0 && thisSum > 0) || (thisSum > 0 && thisSum < minPosSum))
                minPosSum = thisSum;
        }
    }

    return minPosSum;
}


// c 最大子序列乘积， 
int maxProduct(const vector<int>& nums)
{
    int maxpro = nums[0], minpro = nums[0], answer = nums[0];

    for (size_t i = 1; i < nums.size(); ++i)
    {
        int mx = maxpro, mn = minpro;

        maxpro = max(mx * nums[i], max(mn * nums[i], nums[i]));
        minpro = min(mx * nums[i], min(mn * nums[i], nums[i]));
        answer = max(maxpro, answer);
    }
    return answer;
}

leetcode 最大子序列乘积

2.18 折半查找求零点

#include <iostream>
#include <vector>

using namespace std;

double func(double x)
{
    return x * x - x - 1;
}

double solveZero(double (*func)(double x), double low, double high)
{
    while (func(low) * func(high) < 0)
    {
        double mid = (low + high) / 2;

        if (func(mid) == 0)
            return mid;
        else if (func(mid) * func(low) < 0)
            high = mid;
        else if (func(mid) * func(high) < 0)
            low = mid;
    }
}


int main()
{

    cout << solveZero(func, 1, 2) << endl;

    system("pause");
    return 0;
}

2.19 返回最大子序列对应的下标

// 只做了算法4的，算法4有些错误，不能对纯负数进行求解，下面代码已经更正
#include <iostream>
#include <vector>

using namespace std;


int maxSubSum4(const vector<int>& a, int& start, int& end)
{
    start = end = 0;

    int maxSum = a[0], thisSum = 0;  // 原代码maxSum = 0；
    int thisSumStart = 0;

    for (int j = 0; j < a.size(); ++j)
    {
        thisSum += a[j];

        if (thisSum > maxSum)
        {
            maxSum = thisSum;

            start = thisSumStart;
            end = j;
        }
            
        if (thisSum < 0)  // 原代码 else if (thisSum < 0)
        {
            thisSum = 0;
            thisSumStart = j + 1;
        }
            
    }

    return maxSum;
}

int main()
{
    vector<int> vec{ -4, 2, -1, 4, -3, 2, 2 };
    int start, end;

    cout << maxSubSum4(vec, start, end) << endl;
    cout << "(" << start << "," << end << ")" << endl;

    system("pause");
    return 0;
}

2.20

a. 编写一个程序来确定正整数N是否是素数

#include <iostream>
#include <vector>

using namespace std;

bool isPrimeNum(int N)
{
    if (N == 1)
        return false;
    else if (N == 2)
        return true;
    
    for (int i = 2; i <= pow(N, 0.5); ++i)
    {
        if (N % i == 0)
            return false;
    }

    return true;
}

int main()
{
    cout << isPrimeNum(97) << endl;

    system("pause");
    return 0;
}

2.21 输出(0, N]之间的全部素数

// 下面的代码是在解第20题的时候写的，有需要的再修改一下吧

#include <iostream>
#include <vector>

using namespace std;

bool isPrimeNum(int N, vector<int>& vec)
{
    if (N <= 1)
        return false;
    else if (N == 2)
    {
        vec.push_back(2);
        return true; 
    }
        
    vec.push_back(2);
    int n = N;
    for (int i = 3; i <= N; ++i)
    {
        size_t j = 0;
        while (j < vec.size() && vec[j] <= pow(N, 0.5))
            if (i % vec[j++] == 0)
                break;

        if (j >= vec.size() || vec[j] > pow(N, 0.5))
            vec.push_back(i);
    }

    if (*(vec.end() - 1) == N)
        return true;
    return false;
}

int main()
{
    vector<int> primeNumVec;
    cout << isPrimeNum(98, primeNumVec) << endl;

    for (auto iter = primeNumVec.begin(); iter < primeNumVec.end(); ++iter)
    {
        cout << *iter << " ";
    }
    cout << endl;

    system("pause");
    return 0;
}

2.23 不用递归，写出快速求幂的程序

double Pow2(double x, unsigned int n)
{
	double result = 1;
	while (n)
	{
		if (n & 1) 
			result *= x;
		n >>= 1;
		x *= x;
	}
	return result;
}

原文链接

2.26

求主元素,这题有大佬会吗

2.27 矩阵从左到右递增，从上到下递增，确定X是否在矩阵中

#include <iostream>
#include <list>
#include <vector>

using namespace std;

bool existInMatrix(const int elems, const vector<vector<int>>& mat)
{
    size_t curRow = 0;
    size_t curCol = mat[0].size() - 1;

    while (mat[curRow][curCol] != elems)
    {
        if (mat[curRow][curCol] > elems)
        {
            --curCol;
            if (curCol < 0)
                return false;
        }
            
        else
        {
            ++curRow;
            if (curRow >= mat[0].size())
                return false;
        }      
    }

    return true;
}
   

int main()
{

    vector<vector<int>> Mat{ {1, 5, 9, 13},{2, 6, 10, 14}, {3, 7, 11, 15}, {4, 8, 12,16} };
    cout << existInMatrix(7, Mat) << endl;

    system("pause");
    return 0;
}