Missing Number Leetcode 268 Java

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.

Note:

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

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

这道题目和其他题目不同的地方在于，所有的备选数是已知的。所以，切入点就在这。我的第一想法是，将数组中所有的数加起来得到sum1，然后求1，2，3。。。n的和，得到sunm2，然后用sum2-sum1，AP程序如下：

public class Solution {
public int missingNumber(int[] nums) {
if(nums==null||nums.length==0)
return 0;
int len=nums.length;
if(len==1){
return nums[0]==0?1:0;
}
long sum=0;
for(int i=0;i<len;i++){
sum+=nums[i];
}
return (int)(((len)*(len+1))/2-sum);
}
}

AP之后去看了眼讨论区，然后发现，还有更好的解法，切入点是相同的，但是用的不是求和的方法，而是采用了亦或的方式来做。AP程序如下：

public int missingNumber(int[] nums) {
int sum = 0;
for (int i = 0; i <= nums.length; ++i) sum ^= i;
for (int i = 0; i < nums.length; ++i) sum ^= nums[i];
return sum;
}