Leetcode 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：

1.先将数组排序，再遍历一遍即可，但是比较慢

2.（最快的）新建一个数组，数组是原数组长度加一，每个新数组元素初始化为零。遍历原数组，将新数组相应位置的元素设为-1。遍历新数组，找到元素值不为-1的位置，返回该位置索引值。

public class S268 {
    public int missingNumber(int[] nums) {
        //slow 排序会影响时间
/*        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if(i != nums[i])
                return i;
        }
        return nums[nums.length-1]+1;
        */
        //best
        int []nums2 = new int[nums.length+1];
        for (int i = 0; i < nums.length; i++) {
            nums2[nums[i]] = -1;
        }
        int i = 0;
        for (; i < nums2.length; i++) {
            if (nums2[i] != -1)
                break;
        }
        return i;
    }
}