Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.
Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
Example 1:
Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
Example 2:
Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
Example 3:
Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
Example 4:
Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.
Constraints:
- n == nums.length
- 1 <= n <= 104
- 0 <= nums[i] <= n
- All the numbers of nums are unique.
解题思路：
方法1：因为是从0-n的数组，所以可以先用求和公式(n+1)*n/2计算出原本数组和的大小，然后再遍历一遍数组求数组和，用原本数组和-遍历数组和=缺失数字。
方法2：^符求和，xor的特性a^a=0, 0^a=a, 所以通过数组依次遍历循环的index即可得到答案。
class Solution {
public int missingNumber(int[] nums) {
if (nums.length == 0) {
return 1;
}
int lack = nums.length; //因为缺少元素，所以实际长度即是数组里的n
for (int i=0; i< nums.length; i++) {
lack ^= i^nums[i];
}
return lack;
}
}