268. Missing Number （binary）

0到n的sum减去已经存在的就是missing number

 1 //Old
 2 class Solution {
 3     public int missingNumber(int[] nums) {
 4         int max = nums.length;
 5         List<Integer> A = new ArrayList<Integer>();
 6         for(int i = 0; i <= max; i++) {
 7             A.add(i);
 8         }
 9         for(int j = 0; j < nums.length; j++) {
10             A.set(nums[j], -1);
11         }
12         for(Integer a : A) {
13             if( a != -1)
14                 return a;
15         }
16         return 0;
17     }
18 }
19 
20 
21 
22 
23 //New 100%
24 
25 class Solution {
26     public int missingNumber(int[] nums) {
27         int n = nums.length;
28         int sum = (1 + n)* n / 2;
29         for(int a : nums) {
30             sum -= a;
31         }
32         return sum;
33         
34     }
35 }