Binary Search

Question

You are given a sequence of n integers S and a sequence of different q integers T. Write a program which outputs C, the number of integers in T which are also in the set S.

Input In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers are given.

Output Print C in a line.

Constraints n ≤ 10000 q ≤ 500 0 ≤ an element in S ≤ 109 0 ≤ an element in T ≤ 109 Sample Input 1 5 1 2 3 4 5 3 3 4 1 Sample Output 1 3 Sample Input 2 3 3 1 2 1 5 Sample Output 2 0 Sample Input 3 5 1 1 2 2 3 2 1 2 Sample Output 3 2

Meaning

使用二分搜索，效率就会很大提高。但是需要注意，二分搜索的前提是数组已经排序好了。那么排序一般最高的效率希尔排序也是0(n1.25），岂不是效率还降低了？相对于O（n），这里题目限制说了 给出的数据就是按升序排列的，那么时间复杂度就是0（qlogn）

Coding

#include<iostream>
#include<cstdio>
using namespace std;
int A[100005];
int BinarySearch(int a[], int size, int key) {
	int l = 0;
	int r = size - 1;
	while (l <= r) {
		int mid = l + (r - l) / 2;
		if (key == a[mid])
			return true;
		else if (key > a[mid])
			l = mid + 1;
		else
			r = mid - 1;
	}
	return false;//没有找到
}
int main() {
	int n, q, key, sum = 0;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%d", &A[i]);
	scanf("%d", &q);
	for (int i = 0; i < q; i++) {
		scanf("%d", &key);
		if (BinarySearch(A,n,key))
			sum++;
	}
	printf("%d\n", sum);
}

Summary

二分搜索使用前需要对数组进行排序； 标准二分搜索算法：

int BinarySearch(int a[], int size ,int p){
	int L=0; //查找区间的左端点
	int R=size -1;//查找区间的的右端点
	while(L<=R){
		int mid =L+(R-L)/2;
		if(p == a[mid])
			return mid;
		else if(p>a[mid])
			L=mid +1;
		else
			R=mid-1; 
	}
	return -1;
}

废江博客 , 版权所有丨如未注明 , 均为原创丨本网站采用BY-NC-SA协议进行授权 转载请注明原文链接：Binary Search