欢迎访问我的CCF认证解题目录

题目描述

思路过程 

 n最大值为1000，一看就想直接暴力解法，两重for循环，一个一个判断就可以了，时间复杂度为O()

O()代码

import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] number = new int[n];
		int count = 0;
		for ( int i = 0; i < number.length; i++ ) {
			number[i] = in.nextInt();
		}
		for ( int i = 0; i < number.length; i++ ) {
			for ( int j = i+1; j < number.length; j++ ) {
				if (  Math.abs((int)(number[i]-number[j]))== 1 ) {//差值为1
					count++;
				}
			}
		}
		System.out.println(count);
	}
}

我们也可以定义一个数组存储出现的次数，这样i和i-1差值的个数直接用nums[i]*nums[i+1]就可以算出，时间复杂度为O(n)

O(n)代码

import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] nums = new int[10005];
		int sum = 0;
		for ( int i = 0; i < n; i++ ) nums[in.nextInt()]++;
		for ( int i = 0; i < nums.length-1; i++ ) sum += nums[i]*nums[i+1];
		System.out.println(sum);
	}
	
}