C语言练习9---数组查找元素

一个二维数组按从小到大排列，我们可以按照分块查询的方式查找此元素，找到了返回其下标

(1) 普通版

#include <stdio.h>
#include <stdlib.h>

int FindValue(int arr[][5], int abs, int ord, int val)
{
	int i = 0;
	while (val > arr[i][ord - 1] && i<abs-1)    //寻找val所在的行
	{
		i++;
	}

	if (val == arr[i][ord - 1])                 //恰好val==行末
	{
		return i*ord + ord-1;                  
	}


	int j = ord - 1;
	while (j>=0 && arr[i][j]>val)                //在所在行进行线性查找 
	{
		j--;
	}
	if (val > arr[i][j])                         //大于最后一行的末尾元素，说明不存在此数
	{
		return -1;
	}
	if (j < 0)                                 //没找到 
	{
		return -1;
	}
	else                                       //找到了
	{
		return i*ord + j;
	}
}
 
void Judge(int result, int ord)
{
	if (result == -1)
	{
		printf("对不起，数组中没有这个数!\n");
	}
	else
	{
		printf("此数所在的位置:横坐标=%d ，纵坐标=%d \n", result / ord + 1, result % ord + 1);
	}
}

int main()
{
	int arr[4][5] = 
	{
		{ 1, 3, 5, 7, 9}, 
		{11,13,15,17,19}, 
		{21,23,25,27,29}, 
		{31,33,35,37,39}, 
	};
	int val = 25;
	Judge(FindValue(arr, 4, 5, val),5);
	system("pause");
	return 0;
}

(2) 精简版

#include <stdio.h>
#include <stdlib.h>
int FindValue(int arr[][5], int abs, int ord, int val)
{
	int i = 0;
	while (val > arr[i][ord - 1] && i<abs)    
	{
		i++;
	}
	if (i == abs)
	{
		return -1;
	}

	int j = ord - 1;
	while (j >= 0 && arr[i][j]>val)                   
	{
		j--;
	}

	if (j >= 0 && arr[i][j] == val)
	{
		return i*ord + j;
	}
	else
	{
		return -1;
	}
}

void Judge(int result, int ord)
{
	if (result == -1)
	{
		printf("对不起，数组中没有这个数!\n");
	}
	else
	{
		printf("此数所在的位置:横坐标=%d ，纵坐标=%d \n", result / ord + 1, result % ord + 1);
	}
}

int main()
{
	int arr[4][5] = 
	{
		{ 1, 3, 5, 7, 9}, 
		{11,13,15,17,19}, 
		{21,23,25,27,29}, 
		{31,33,35,37,39}, 
	};
	int val = 35;
	Judge(FindValue(arr, 4, 5, val),5);
	system("pause");
	return 0;
}

有时候正向思维的确比逆向思维能简单不少~~~

二分查询

#include <stdio.h>
#include <stdlib.h>

int Binary_Que(int *cp, int size,int val)
{
	int left = 0;
	int right = size - 1;
	int pos = -1;
	int mid = 0;
	while (left <= right)
	{
		mid = (left + right) / 2;
		if (cp[mid] == val)
		{
			pos = mid;
			break;            //返回下标
		}
		else if (cp[mid] < val)
		{
			left = mid + 1;
		}
		else
		{
			right = mid - 1;
		}
	}
	return pos;
}

int main()
{
	int arr[] = { 12, 23, 34, 45, 56, 78, 89, 90, 100, 110, 120 };
	int size = sizeof(arr) / sizeof(arr[0]);
	int val = 100;
	int accepval = Binary_Que(arr, size, val);
	system("pause");
	return 0;
}

重复数据找到起始元素下标

线性探测

#include <stdio.h>
#include <stdlib.h>

int Binary_Que(int *cp, int size,int val)
{
	int left = 0;
	int right = size - 1;
	int pos = -1;
	int mid = 0;
	while (left <= right)
	{
		mid = (left + right) / 2;
		if (cp[mid] == val)
		{
			
			while (cp[mid] == val)
			{
				pos = mid;
				mid--;
			}
			break;
		}
		else if (cp[mid] < val)
		{
			left = mid + 1;
		}
		else
		{
			right = mid - 1;
		}
	}
	return pos;
}

int main()
{
	int arr[] = { 12, 12, 12, 12, 12, 12, 12, 23, 23, 23, 23, 23, 34, 45, 56, 78, 89, 90, 100, 110, 120 };
	int size = sizeof(arr) / sizeof(arr[0]);
	int val = 100;
	int accepval = Binary_Que(arr, size, val);
	system("pause");
	return 0;
}

若有连续相同的数出现，在若干个相同的数找第一个数的下标，可再用二分法(更快速)

#include <stdio.h>
#include <stdlib.h>

int Binary_Que(int *cp, int size,int val)
{
	int left = 0;
	int right = size - 1;
	int pos = -1;
	int mid = 0;
	while (left <= right)
	{
		mid = (left + right) / 2;
		if (cp[mid] == val)
		{
			if (mid > left&&cp[mid - 1] == val)
			{
				right = mid - 1;
			}
			else
			{
				pos = mid;
				break;
			}
			
		}
		else if (cp[mid] < val)
		{
			left = mid + 1;
		}
		else
		{
			right = mid - 1;
		}
	}
	return pos;
}

int main()
{
	int arr[] = { 12, 12, 12, 12, 12, 12, 12, 23, 23, 23, 23, 23, 34, 45, 56, 78, 89, 90, 100, 110, 120 };
	int size = sizeof(arr) / sizeof(arr[0]);
	int val = 23;
	int accepval = Binary_Que(arr, size, val);
	system("pause");
	return 0;
}