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
     
       A = new ArrayList
      
       (); 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 }