【Leetcode】Missing Number

题目链接：https://leetcode.com/problems/missing-number/

题目：

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?

思路：

1、数组是0~n，所以可以利用数组和

2、用异或操作，让nums数组和0~n异或，重复的元素异或结果为0，最后结果就是miss number

算法1：

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

算法2：

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