268. Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example 1:

Input: [3,0,1]
Output: 2

Example 2:

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

//Time: O(n), Space: O(1)
//Approach 1 : 用前n+1项和公式减去nums里面的每一个值，剩下的就是missing number

public int missingNumber(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int result = 0;
        
        for (int i = 0; i < nums.length; i++) {
            result += nums[i];
        }
        
        return nums.length * (nums.length + 1) / 2 - result;

//Approach 2: XOR
    public int missingNumber(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int result = 0;
        for (int i = 0; i < nums.length; i++) {
            result = result ^ i ^ nums[i];
        }
        
        return result ^ nums.length;//最后不要忘记异或最后一个数
    }
//Approach 3: 如果是排序好的数组，可以用二分法time为O(logn), 否则排序time是O(nlogn)
代码略