leetcode268~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?

推荐 异或 方法

public class MissingNumber {    public int missingNumber2(int[] nums) {        if(nums==null || nums.length==0) return 0;        Arrays.sort(nums);        int number;        for(number=0;number<nums.length;number++) {            if(nums[number]!=number) {                return number;            }        }        //如果是{1,2,3}则返回4        return number;    }    //二分法    public int missingNumber3(int[] nums) { //binary search        Arrays.sort(nums);        int left = 0, right = nums.length, mid= (left + right)/2;        while(left<right){            mid = (left + right)/2;            if(nums[mid]>mid) right = mid; //说明左边有遗漏的数            else left = mid+1;        }        return left;    }    //异或  两个相同的数异或为0 则把数组中的每个数和下标全部异或一次，最后的那个数肯定是没出现的数    public int missingNumber(int[] nums) {        int res = nums.length;        for(int i=0;i<nums.length;i++) {            res ^= i;            res ^= nums[i];        }        return res;    }}