LeetCode & Q268-Missing Number-Easy

Array Math Bit Manipulation

Description:

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.

这题真是一言难尽....刚开始不太理解，数字要从0开始往后记，我本来写的Binary Search有问题...参考了Discuss里重写了一遍

看到Discuss里很多用XOR写的，以前没有涉及过这个领域，感觉好神奇...

my Solution:

public class Solution {
    public int missingNumber(int[] nums) {
        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;
    }
}

XOR Solution:

// a^b^b = a, nums[index] = index
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;
}

看到最快的是用数学稍微转换下思路的方法，数学好的人真厉害...其实也不是特别厉害的数学....可是我怎么就没想起来呢...

public class Solution {
    public int missingNumber(int[] nums) {

        int n = nums.length;
        int sum = (n*(n+1)) /2;

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

        int missingNumber = sum - numSum;
        return missingNumber;
    }
}