数据结构与算法分析——C++语言描述(第四版)Mark Allen Weiss 习题 第二章 算法分析
作者:互联网
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;
}
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;
}
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;
}
标签:return,int,C++,high,算法,low,习题,include,thisSum 来源: https://blog.csdn.net/LiuYun023/article/details/121994095