Missing Number Leetcode 268 Java__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.

這道題目和其他題目不同的地方在於，所有的備選數是已知的。所以，切入點就在這。我的第一想法是，將數組中所有的數加起來得到sum1，然後求1，2，3。。。n的和，得到sunm2，然後用sum2-sum1，AP程式如下：

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之後去看了眼討論區，然後發現，還有更好的解法，切入點是相同的，但是用的不是求和的方法，而是採用了亦或的方式來做。AP程式如下：

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;}