//★题目:在行和列都排好序的矩阵中找数
//要求:给定一个有N×M的整型矩阵matrix和一个整数K,matrix的每一行每一列都是排好序的。
// 实现一个函数,判断K是否在matrix中。要求时间复杂度为O(N+M),空间复杂度为O(1)
// 例如:0 1 2 5
// 2 3 4 7
// 4 4 4 8
// 5 7 7 9
// 如果K = 7,返回true;如果K = 6,返回false。
#include <iostream>
#include <vector>
using namespace std;
bool isContains(vector<vector<int>> matrix, int k);
void printVector2(vector<vector<int>> matrix);
vector<vector<int>> generateDesignated2Vector(int *arr, int setCols, int setRows);//生成二维矩阵
int main()
{
//int *arr = new int[10];
int arrayMine[16] = { 0, 1, 2, 5, 2, 3, 4, 7, 4, 4, 4, 8, 5, 7, 7, 9 };
int *arr = arrayMine;
vector<vector<int>> matrix = generateDesignated2Vector(arr, 4, 4);
printVector2(matrix);
cout << endl;
int testNum = 7;
bool isConatain = isContains(matrix, testNum);
if (isConatain) cout << "true";
else cout << "flase";
//printMatrixZigZag(matrix);
system("pause");
return 0;
}
bool isContains(vector<vector<int>> matrix, int k)
{
int col = 0;
int row = matrix.size()-1;
//cout << matrix[row][col];
while (col<matrix[0].size())
{
if (k==matrix[row][col])
{
return true;
}
else if (k>matrix[row][col])
{
col++;
}
else if (k<matrix[row][col])
{
while (row>=0)
{
if (k == matrix[row][col]) return true;
row--;
}
col++;
row = matrix.size() - 1;
}
}
return false;
}
void printVector2(vector<vector<int>> matrix)//打印二维矩阵
{
int cols = matrix[0].size();
int rows = matrix.size();
for (int i = 0; i < rows; i++)
{
for (int j = 0; j < cols; j++)
{
cout << matrix[i][j];
if (matrix[i][j]<10) cout << " ";
else cout << " ";
}
cout << endl;
}
}
vector<vector<int>> generateDesignated2Vector(int *arr, int setCols, int setRows)//生成二维矩阵
{
int setNum = 0;
int arrIndex = 0;
vector<vector<int>> result;
vector<int> tmp;
if ((setCols == 0) || (setRows == 0))
{
return result;
}
static int rows = 0;
while (rows < setRows)
{
tmp.clear();
for (int cols = 0; cols < setCols; cols++)
{
setNum = arr[arrIndex];
arrIndex++;
tmp.push_back(setNum);
}
result.push_back(tmp);
rows++;
}
return result;
}
数组与矩阵:在行和列都排好序的矩阵中找数
最新推荐文章于 2019-09-02 15:41:59 发布