【PAT甲级 LinkedHashMap】1041 Be Unique (20 分) Java、C++版

题目

题目的意思是，找到第一个不重复的元素。输出它。

一开始用Java做，试了好多种方法都超时。后来换用C++通过了。

AC题解：C++版

用Java尝试了各种方式，都是后两个测试点过不去，只好换C++

这个算法挺巧妙的，注意记录输入顺序的方式

#include <iostream>
using namespace std;
int main()
{int    num[100001], count[10001];int   n;cin >> n;for ( int i = 0; i < n; i++ ){cin >> num[i];count[num[i]]++;}for ( int i = 0; i < n; i++ ){if ( count[num[i]] == 1 ){cout << num[i];return(0);}}cout << "None";return(0);
}

超时解法1：使用LinkedHashMap

提问：java中如何统计数组中出现相同元素的个数？

回答：你可以用map类型，思路大概是这样的：

把数组从第0个开始保存在<key,value>里
作一个对比过程，其中key表示元素，存的过程作一个对比，如果相同，则其相应的value值+1
最后输出key：value

import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int total = sc.nextInt();// 使用map存储Map<Integer, Integer> map = new LinkedHashMap<Integer, Integer>();for (int i = 0; i < total; i++) {int num = sc.nextInt();Integer v = map.get(num);if (v != null) {map.put(num, v + 1);} else {map.put(num, 1);}}// 遍历mapIterator<?> iter = map.entrySet().iterator();while (iter.hasNext()) {Map.Entry<Integer, Integer> entry = (Entry<Integer, Integer>) iter.next();if ((Integer) entry.getValue() == 1) {System.out.println(entry.getKey());return;}}System.out.println("None");}
}

超时解法2：同样的算法，C++不超时

读取数据时，BufferedReader比使用Scanner快一些
使用Scanner要花120多ms，BufferedReader大约80ms

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;public class Main {public static void main(String[] args) throws IOException {BufferedReader in = new BufferedReader(new InputStreamReader(System.in));String aa = in.readLine();String[] arr = aa.split(" ");int n = Integer.parseInt(arr[0]);int[] num = new int[100001];int[] count = new int[100001];for (int i = 0; i < n; i++) {num[i] = Integer.parseInt(arr[i + 1]);count[num[i]]++;}for (int i = 0; i < n; i++) {if (count[num[i]] == 1) {System.out.println(num[i]);return;}}System.out.println("None");}
}

使用Scanner代替BufferedReader：超时更严重

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();int[] num = new int[100001];int[] count = new int[100001];for (int i = 0; i < n; i++) {num[i] = sc.nextInt();count[num[i]]++;}for (int i = 0; i < n; i++) {if (count[num[i]] == 1) {System.out.println(num[i]);return;}}System.out.println("None");}
}

超时解法3：两层for循环

暴力O(n^2)循环嵌套，耗时也不过如此嘛，虽然超时了，但是居然比Map还快一点点

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int total = sc.nextInt();int[] arr = new int[total];for (int i = 0; i < total; i++) {arr[i] = sc.nextInt();}// 找第一个不重复的int i, j;for (i = 0; i < total; i++) {for (j = 0; j < total; j++) {if(i==j)continue;if (arr[i] == arr[j]) {break;}}if (j == total) {System.out.println(arr[i]);return;}}System.out.println("None");}
}

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