题目要求时间复杂度小于O(N)
#include<stdio.h>
int find(int arr[][3], int *px, int *py,int key)
{
int x = 0;
int y = *py - 1;
while ((x < *px) && (y >= 0))
{
if (arr[x][y] == key)
{
//下标由指针带回
*px = x;
*py = y;
return 0;
}
else if (arr[x][y] < key)
{
x++;
}
else
{
x++;
y--;
}
}
*px = -1;
*py = -1;
}
void print_Arrs(int *arr,int row, int col)//输出函数（指针法）得益于c语言线性存储
{
int i = 0;
int j = 0;
for (i = 0; i < row; i++)
{
for (j = 0; j < col; j++)
{
printf("%d ", arr[i*col + j]);
}
printf("\n");
}
}
int main()
{
int n;
int arr[3][3] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
print_Arrs((int *)arr, 3, 3);
int x = 3;
int y = 3;
printf("请输入需要查找的数字:\n");
scanf("%d", &n);
find(arr,&x,&y,n);
if (x != -1 && y != -1)
{
printf("找到了，下标为%d,%d\n",x,y);
}
else
{
printf("没找到！\n");
}
return 0;
}
