README_EN.md

17.19. Missing Two

中文文档

Description

You are given an array with all the numbers from 1 to N appearing exactly once, except for two number that is missing. How can you find the missing number in O(N) time and 0(1) space?

You can return the missing numbers in any order.

Example 1:

Input: [1]

Output: [2,3]

Example 2:

Input: [2,3]

Output: [1,4]

Note:

• nums.length <= 30000

Solutions

Solution 1

class Solution:
    def missingTwo(self, nums: List[int]) -> List[int]:
        n = len(nums) + 2
        xor = 0
        for v in nums:
            xor ^= v
        for i in range(1, n + 1):
            xor ^= i

        diff = xor & (-xor)
        a = 0
        for v in nums:
            if v & diff:
                a ^= v
        for i in range(1, n + 1):
            if i & diff:
                a ^= i
        b = xor ^ a
        return [a, b]

class Solution {
    public int[] missingTwo(int[] nums) {
        int n = nums.length + 2;
        int xor = 0;
        for (int v : nums) {
            xor ^= v;
        }
        for (int i = 1; i <= n; ++i) {
            xor ^= i;
        }
        int diff = xor & (-xor);
        int a = 0;
        for (int v : nums) {
            if ((v & diff) != 0) {
                a ^= v;
            }
        }
        for (int i = 1; i <= n; ++i) {
            if ((i & diff) != 0) {
                a ^= i;
            }
        }
        int b = xor ^ a;
        return new int[] {a, b};
    }
}

class Solution {
public:
    vector<int> missingTwo(vector<int>& nums) {
        int n = nums.size() + 2;
        int eor = 0;
        for (int v : nums) eor ^= v;
        for (int i = 1; i <= n; ++i) eor ^= i;

        int diff = eor & -eor;
        int a = 0;
        for (int v : nums)
            if (v & diff) a ^= v;
        for (int i = 1; i <= n; ++i)
            if (i & diff) a ^= i;
        int b = eor ^ a;
        return {a, b};
    }
};

func missingTwo(nums []int) []int {
	n := len(nums) + 2
	xor := 0
	for _, v := range nums {
		xor ^= v
	}
	for i := 1; i <= n; i++ {
		xor ^= i
	}
	diff := xor & -xor
	a := 0
	for _, v := range nums {
		if (v & diff) != 0 {
			a ^= v
		}
	}
	for i := 1; i <= n; i++ {
		if (i & diff) != 0 {
			a ^= i
		}
	}
	b := xor ^ a
	return []int{a, b}
}