README_EN.md

645. Set Mismatch

中文文档

Description

You have a set of integers s, which originally contains all the numbers from 1 to n. Unfortunately, due to some error, one of the numbers in s got duplicated to another number in the set, which results in repetition of one number and loss of another number.

You are given an integer array nums representing the data status of this set after the error.

Find the number that occurs twice and the number that is missing and return them in the form of an array.

 

Example 1:

Input: nums = [1,2,2,4]
Output: [2,3]

Example 2:

Input: nums = [1,1]
Output: [1,2]

 

Constraints:

• 2 <= nums.length <= 104
• 1 <= nums[i] <= 104

Solutions

Python3

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        eor, n = 0, len(nums)
        for i in range(1, n + 1):
            eor ^= (i ^ nums[i - 1])
        diff = eor & (~eor + 1)
        a = 0
        for i in range(1, n + 1):
            if (nums[i - 1] & diff) == 0:
                a ^= nums[i - 1]
            if (i & diff) == 0:
                a ^= i
        b = eor ^ a
        for num in nums:
            if a == num:
                return [a, b]
        return [b, a]

Java

class Solution {
    public int[] findErrorNums(int[] nums) {
        int eor = 0;
        for (int i = 1; i <= nums.length; ++i) {
            eor ^= (i ^ nums[i - 1]);
        }
        int diff = eor & (~eor + 1);
        int a = 0;
        for (int i = 1; i <= nums.length; ++i) {
            if ((nums[i - 1] & diff) == 0) {
                a ^= nums[i - 1];
            }
            if ((i & diff) == 0) {
                a ^= i;
            }
        }
        int b = eor ^ a;
        for (int num : nums) {
            if (a == num) {
                return new int[]{a, b};
            }
        }
        return new int[]{b, a};
    }
}

TypeScript

function findErrorNums(nums: number[]): number[] {
    let xor = 0;
    for (let i = 0; i < nums.length; ++i) {
        xor ^= (i + 1) ^ nums[i];
    }

    let divide = 1;
    while ((xor & divide) == 0) {
        divide <<= 1;
    }

    let ans1 = 0,
        ans2 = 0;
    for (let i = 0; i < nums.length; ++i) {
        let cur = nums[i];
        if (divide & cur) {
            ans1 ^= cur;
        } else {
            ans2 ^= cur;
        }

        let idx = i + 1;
        if (divide & idx) {
            ans1 ^= idx;
        } else {
            ans2 ^= idx;
        }
    }
    return nums.includes(ans1) ? [ans1, ans2] : [ans2, ans1];
}

C++

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        int eor = 0, n = nums.size();
        for (int i = 1; i <= n; ++i) {
            eor ^= (i ^ nums[i - 1]);
        }
        int diff = eor & (~eor + 1);
        int a = 0;
        for (int i = 1; i <= n; ++i) {
            if ((nums[i - 1] & diff) == 0) {
                a ^= nums[i - 1];
            }
            if ((i & diff) == 0) {
                a ^= i;
            }
        }
        int b = eor ^ a;
        for (int num : nums) {
            if (a == num) {
                return {a, b};
            }
        }
        return {b, a};
    }
};

Go

把每个数都放到它应该在的位置，最后出现“异常”的就是重复的数和丢失的数。

func findErrorNums(nums []int) []int {
	n := len(nums)
	for i := 0; i < n; i++ {
		for nums[i] != i+1 && nums[nums[i]-1] != nums[i] {
			nums[i], nums[nums[i]-1] = nums[nums[i]-1], nums[i]
		}
	}
	for i := 0; i < n; i++ {
		if nums[i] != i+1 {
			return []int{nums[i], i + 1}
		}
	}
	return []int{-1, -1}
}

也可以使用位运算。

func findErrorNums(nums []int) []int {
	eor, n := 0, len(nums)
	for i := 1; i <= n; i++ {
		eor ^= (i ^ nums[i-1])
	}
	diff := eor & (-eor)
	a := 0
	for i := 1; i <= n; i++ {
		if (nums[i-1] & diff) == 0 {
			a ^= nums[i-1]
		}
		if (i & diff) == 0 {
			a ^= i
		}
	}
	b := eor ^ a
	for _, num := range nums {
		if a == num {
			return []int{a, b}
		}
	}
	return []int{b, a}
}

C++

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        int eor = 0, n = nums.size();
        for (int i = 1; i <= n; ++i) {
            eor ^= (i ^ nums[i - 1]);
        }
        int diff = eor & (~eor + 1);
        int a = 0;
        for (int i = 1; i <= n; ++i) {
            if ((nums[i - 1] & diff) == 0) {
                a ^= nums[i - 1];
            }
            if ((i & diff) == 0) {
                a ^= i;
            }
        }
        int b = eor ^ a;
        for (int num : nums) {
            if (a == num) {
                return {a, b};
            }
        }
        return {b, a};
    }
};

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