268. Missing Number

Description

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example 1

Input: [3,0,1]
Output: 2
Example 2

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

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

Solution

有趣的题目。最直观的想法是用sum去做，虽然能过，但如果n非常大的话，sum是会overflow的。
更好的办法是用XOR，利用a ^ a ^ b == b的原理。
如果数组有序，用二分最优。

XOR, time O(n), space O(1)

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