Questions and tests
package sword040;
/*一个整型数组里除了两个数字之外，其他的数字都出现了两次。请写程序找出这两个只出现一次的数字。
*/
public class main {
public static void main(String[] args) {
int[][] testTable = {
{1,2},{1,1,2,2,5,6,7,7},{1,2,3,3},{1,1,2,3}};
for (int[] ito : testTable) {
test(ito);
}
}
private static void test(int[] ito) {
Solution solution = new Solution();
int rtn;
long begin = System.currentTimeMillis();
for (int i = 0; i < ito.length; i++) {
System.out.print(ito[i]+" ");
}//开始时打印数组
int[] num1 = new int[1];
int[] num2 = new int[1];
solution.findNumsAppearOnce(ito,num1,num2);//执行程序
long end = System.currentTimeMillis();
System.out.println("rtn1=" + num1[0]);
System.out.println("rtn2=" + num2[0]);
System.out.println();
System.out.println("耗时：" + (end - begin) + "ms");
System.out.println("-------------------");
}
}
Solution 1 (didn't figure it out)
Using the property that two identical numbers are XORed to 0, the binary form of the XOR operation result of two non-zero and different numbers must have the property of
1. 1. XOR the array from left to right to get a value of
2. Calculate the position index of the lowest bit 1 of the binary value.
3. Divide the original array into two groups according to whether the index is 1 or 0. The division of the two groups is a logical division, and each group is continuously XORed to get two numbers that appear once.
//num1,num2分别为长度为1的数组。传出参数
//将num1[0],num2[0]设置为返回结果
public class Solution {
public void FindNumsAppearOnce(int [] array,int num1[] , int num2[]) {
int res=0;
for(int i=0;i<array.length;i++){
res ^=array[i];
}
int index=0;
while(true){//有的代码是for循环，且循环次数i<32；我不懂为什么只循环32次？
if((res>>index&1)==1){
break;
}
index++;
}
for (int i=0;i<array.length;i++){
if((array[i]>>index &1)==1){
num1[0] ^=array[i];
}
else num2[0] ^=array[i];
}
}
}