1、自己答案
public class Text {
public static void main(String[] args) {
int[] A = new int[]{1,5,8,8,9,2,5,2,3,1,3};
int i = new Text().singleNumber(A);
System.out.println(i);
}
/**
* @param A: An integer array
* @return: An integer
*/
public int singleNumber(int[] A) {
Set<Integer> set = new HashSet<>();
for (int i = 0;i<A.length ;i++ ){
boolean flag = set.add(A[i]);
if (!flag) {
set.remove(A[i]);
}
}
Iterator<Integer> iterator = set.iterator();
return iterator.next();
}
}
更优答案
public class Text {
public static void main(String[] args) {
int[] A = new int[]{1,5,8,8,9,2,5,2,3,1,3};
int i = new Text().singleNumber(A);
System.out.println(i);
}
/**
* @param A: An integer array
* @return: An integer
*/
public int singleNumber(int[] A) {
if(A == null || A.length == 0) {
return -1;
}
int rst = 0;
for (int i = 0; i < A.length; i++) {
rst ^= A[i];
}
return rst;
}
}
解读:异或运算,不同为1,相同为0,任意相同数字返回0(即0000 0000),只有一个单独的数字与0做异或运算,例0000 0101,相同返回0,不同返回1,即返回该数字本身,一次遍历获取结果,时间复杂度O(n)