Title link: Numbers that appear only once in LeeCode136
Title description: The way 
I think is to sort them one by one. If the current one is equal to the previous element, then set the flag to true. When the flag is true, it can go forward even if it is not equal. Go, find the point that is neither flag equal to the previous element nor false is a single element
class Solution {
public int singleNumber(int[] nums) {
Arrays.sort(nums);
boolean flag=false;
int index=1;
while (index < nums.length) {
if(nums[index-1]==nums[index]){
flag=true;
}else {
if(flag){
index++;
flag=false;
continue;
}else{
return nums[index-1];
}
}
index++;
}
return nums[index-1];
}
}
Then the efficiency was 50%, and I looked for a solution to the problem, which was stupid. . . . It will make me think that my life will be wasted
class Solution {
public int singleNumber(int[] nums) {
for (int i = 1; i < nums.length; i++) {
nums[0]^=nums[i];
}
return nums[0];
}
}