LeetCode: 268. Missing Number

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?

Credits: Special thanks to @jianchao.li.fighter for adding this
problem and creating all test cases.

自己的答案，13ms：

public class Solution {
    public int missingNumber(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        Arrays.sort(nums);
        int i = 0;
        for (i = 0; i < nums.length; i++) {
            if (nums[i] != i) {
                return i;
            }
        }
        return i;
    }
}

最快的答案，1ms：

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

        return (nums.length * (nums.length + 1)) / 2 - sum;
    }
}

3 different ideas: XOR, SUM, Binary Search. Java code
xor:

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

求和：

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

二分查找：

public int missingNumber(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;
        else left = mid+1;
    }
    return left;
}