268. Missing Number

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?

思路一：

求和

思路二：

异或运算

The basic idea is to use XOR operation. We all know that a^b^b =a, which means two xor operations with the same number will eliminate the number and reveal the original number.

In this solution, I apply XOR operation to both the index and value of the array. In a complete array with no missing numbers, the index and value should be perfectly corresponding( nums[index] = index), so in a missing array, what left finally is the missing
number.

public int missingNumber(int[] nums) {

int xor = 0, i = 0;
for (i = 0; i < nums.length; i++) {
xor = xor ^ i ^ nums[i];
}

return xor ^ i;
}