public class FindNumsAppearOnce {
//一个整型数组里除了两个数å—之外，其他的数å—都出现了两次。请写程åºæ‰¾å‡ºè¿™ä¸¤ä¸ªåªå‡ºçŽ°ä¸€æ¬¡çš„æ•°å—。
//
//对于ä½è¿ç®—有:b&1=b;b^0=b;
//对于数è¿ç®—：x^0=x;x^x=0;对于b为00001,00010.....当 x&b==bæ—¶å¯å¾—该ä½æ•°ä¸º1;对于全为1的一个数n有 n&x=x;
public static void main(String args[]) {
int array[] = {2, 4, 4, 5, 3, 3, 9, 9};
int num1[] = {0};
int num2[] = {0};
FindNumsAppearOnce.FindNumsAppearOnce(array, num1, num2);
System.out.println(num1[0]);
System.out.println(num2[0]);
}
//todo
//test
public static void FindNumsAppearOnce(int[] array, int num1[], int num2[]) {
///解题æ€è·¯ï¼šæ ¹æ®ä¸€ä¸ªæ•°ä¸Ž0的异或结果为本身，把所有的数进行异或，的到的是ä¸åŒæ•°çš„å¼‚æˆ–ï¼Œæ ¹æ®æ˜“è´§ 的结果的1çš„ä½ç½®è¿›è¡Œåˆ†ç»„
if (array == null || array.length < 2)
return;
int temp = 0;
for (int i = 0; i < array.length; i++)
temp ^= array[i];
int indexOf1 = findFirstBitIs(temp);
for (int i = 0; i < array.length; i++) {
if (isBit(array[i], indexOf1))
num1[0] ^= array[i];
else
num2[0] ^= array[i];
}
}
//00
public static int findFirstBitIs(int num) {
int indexBit = 0;
while (((num & 1) == 0) && (indexBit) < 8 * 4) {
num = num >> 1;
++indexBit;
}
return indexBit;
}
public static boolean isBit(int num, int indexBit) {
num = num >> indexBit;
return (num & 1) == 1;
}
public static void FindNumsAppearO(int [] array,int num1[] , int num2[]){
Integer s=new Integer(6);
int count =0;
int flag=0;
for(int i=0;i