leetcode268: Missing Number

要求：

Given an array containing n distinct numbers taken from 

0, 1, 2, ..., n

,
find the one that is missing from the array.

For example,

Given nums = 

[0, 1, 3]

 return 

2

.

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

注意：将数组放在新数组中，再将数字1-n放入新数组中。将新数组中所有元素进行异或，结果就是那个没有重复的数字，也就是Missing Number

例如：

int[] array={0,1,2,4,5}在新数组中先放入0-5，再放入array中所有元素
int[] newarray={0,1,2,3,4,5,0,1,2,4,5}只有3没有重复两次。
进行异或后，返回单个元素3

public int missingNumber(int[] nums) {
int[] ar = new int[2 * nums.length + 1];
for (int i = 0; i <= nums.length; i++)
ar[i] = i;
for (int i = 0; i < nums.length; i++)
ar[i + nums.length + 1] = nums[i];
int result = ar[0];
for (int i = 1; i < ar.length; i++)
result = result ^ ar[i];
return result;
}