README_EN.md

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difficulty	Hard
edit_url	https://github.com/doocs/leetcode/edit/main/lcci/17.19.Missing%20Two/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

Python3

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]

Java

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};
    }
}

C++

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};
    }
};

Go

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}
}

Swift

class Solution {
    func missingTwo(_ nums: [Int]) -> [Int] {
        let n = nums.count + 2
        var xor = 0

        for num in nums {
            xor ^= num
        }

        for i in 1...n {
            xor ^= i
        }

        let diff = xor & (-xor)

        var a = 0

        for num in nums {
            if (num & diff) != 0 {
                a ^= num
            }
        }

        for i in 1...n {
            if (i & diff) != 0 {
                a ^= i
            }
        }

        let b = xor ^ a
        return [a, b]
    }
}