【刷题】只出现一次的数字(三种解法)

0^0=0   0^1=1   1^0=1  1^1=0

10100011 ^ 00000111 = 10100100

public void swap(int a,int b){
 if (a != b){
        a ^= b;
        b ^= a;
        a ^= b;
    }
}

public int singleNumber (int[] nums) {
   int temp = 0;
       for(int i = 0;i < nums.length;i++){
       temp ^= nums[i];
       }
       return temp;
    }

public int singleNumber(int[] nums) {
        Set<Integer> set = new HashSet<>();
        //遍历数组
        for (int i = 0; i < nums.length; i++) {
            int key = nums[i];
            if (set.contains(key)) {
                set.remove(key);
            } else {
                set.add(key);
            }
        }
        //我们这里采用遍历数组获取目标元素
        for (int i = 0; i < nums.length; i++) {
            int key = nums[i];
            if (set.contains(key)) {
                return key;
            }
        }
        return -1; //没有找到返回-1
    }

public int singleNumber (int[] nums) {
      Map<Integer,Integer> map = new HashMap<>();
      //遍历数组
        for (int i = 0; i < nums.length; i++) {
            int key = nums[i];
           int value =  map.getOrDefault(key,0);
           map.put(key,value+1);
        }
        //迭代Map集合,找到value等于1的key
        for (Map.Entry<Integer,Integer> entrySet:
                map.entrySet()) {
            if (entrySet.getValue()==1){
                return entrySet.getKey();
            }
        }
        return -1;
    }