二维数组中查找 有序数组合并 效率O(n)

新类型矩阵:

1    2    3   4   5

6    7   8   9   10

11 12  13 14 15

16 17  18 19  20

查找一个指定数据是否在矩阵中

#include <iostream>
using namespace std;
bool Find(int* matrix, int row, int colum, int num);

int main()
{
int matrixOne[12] = {1,2,3,4,5,6,7,8,9,10,11,12};
int* matrixTwo = NULL;
bool bl =Find(matrixOne, 3, 4,-3);

if (bl)
{
cout<<"find it"<<endl;
}
else
cout<<"cann't find it"<<endl;
return 0;
}

bool Find(int* matrix, int row, int colum, int num)
{

/////左下角
bool fund = false;
if ((matrix == NULL) || (row <=0 )|| (colum<=0))
{
return fund;
}
if (num < matrix[0] || num >matrix[row*colum -1])
{
return fund;
}
int r = row -1;
int col = 0;
while ((r >=0) &&(col < colum))
{
if (matrix[r*colum +col] == num)
{
fund = true;
cout<<"row = "<< r<<endl;
cout<<"colum =" <<col<<endl;
return fund;
}
else if (matrix[r*colum +col] < num)
col++;
else
r--;
}
return fund;
}

合并两个数组(包括字符串)时,如果从前往后复制每个数字(或字符)需要移动多次,那么我们可以考虑从后往前复制,从而减少移动次数,提高效率

例如:两个连续的数组A1和A2,内存在A1的末尾有足够空间可以容纳A2。请实现一个函数,把A2中的所有数字插入到A1中并且所有的数字是排序的。

#include <iostream>
#include <string>
using namespace std;
void CombineArray(int destr[], int sostr[], int dLen, int sLen);
#define MAX_SIZE 100
#define MIN_SIZE 50

int main()
{
int dlen = 5;
int slen = 4;
int desArray[MAX_SIZE] = {1, 3, 5, 7, 9};
int sorArray[MIN_SIZE] = {2, 4, 6, 8};
CombineArray(desArray, sorArray, dlen, slen );
for (int i=0; i<dlen+slen; i++)
{
cout<<desArray[i]<<" ";
}
cout <<endl;
return 0;
}

void CombineArray(int deArray[], int soArray[], int dLen, int sLen)
{
if ((deArray == NULL) || (soArray == NULL) || (dLen <=0) || (sLen <=0))
{
return;
}
if (MAX_SIZE < dLen + sLen)
{
return;
}
int totalLen = dLen + sLen -1;
int i_spnt = sLen - 1;
int i_dpnt = dLen - 1;
while((i_dpnt>=0) && (i_spnt >=0))
{
if (deArray[i_dpnt] > soArray[i_spnt])
{
deArray[totalLen--] = deArray[i_dpnt--];
}
else
{
deArray[totalLen--] =soArray[i_spnt--];
}
}
while(i_dpnt >=0)
{
deArray[totalLen--] = deArray[i_dpnt--];
}
while(i_spnt >= 0)
{
deArray[totalLen--] =soArray[i_spnt--];
}
return ;
}
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原文地址:https://www.cnblogs.com/niupan369/p/4495442.html