一、内容
Find your present!
In the new year party, everybody will get a "special present".Now it's your turn to get your special present, a lot of presents now putting on the desk, and only one of them will be yours.Each present has a card number on it, and your present's card number will be the one that different from all the others.For example, there are 5 present, and their card numbers are 1, 2, 3, 2, 1.so your present will be the one with the card number of 3, because 3 is the number that different from all the others.
Input
The input file will consist of several cases.
Each case will be presented by an integer n (1<=n<=200, and n is odd) at first. Following that, n positive integers will be given in a line. These numbers indicate the card numbers of the presents.n = 0 ends the input.
Output
For each case, output an integer in a line, which is the card number of your present.
Sample Input
5
1 1 3 2 2
3
1 2 1
0
Sample Output
3
2
二、思路
-
解法一:
异或运算符是^。因为a^a=0,那么要找出单独的数(唯一一个出现一次的数),只需要将所有的数进行异或运算即可。 -
解法二:
使用Map,map<String, Integer> 第一个保存数字,第二个保存次数,由于只会出现一次,遍历map若次数等于1输出即可。
三、代码
解法一:
import java.util.*;
public class HDU_1563 {
static int n, a;
static int ans;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (true) {
n = sc.nextInt();
if (n == 0) return;
ans = 0;
for (int i = 0; i < n; i++) {
a = sc.nextInt();
ans ^= a;
}
System.out.println(ans);
}
}
}
解法二:
import java.util.*;
public class HDU_1563 {
static int n;
static HashMap<String, Integer> map = new HashMap<String, Integer>();
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (true) {
n = sc.nextInt();
if (n == 0) return;
for (int i = 0; i < n; i++) {
String str = sc.next();
if (map.containsKey(str)) {
map.put(str, map.get(str) + 1);
} else {
map.put(str, 1);
}
}
for (String key : map.keySet()) {
if (map.get(key) == 1) {
System.out.println(key);
break;
}
}
map.clear();
}
}
}