README_EN.md

448. Find All Numbers Disappeared in an Array

中文文档

Description

Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.

 

Example 1:

Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]

Example 2:

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

 

Constraints:

• n == nums.length
• 1 <= n <= 105
• 1 <= nums[i] <= n

 

Follow up: Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Solutions

Python3

class Solution:
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        for num in nums:
            idx = abs(num) - 1
            if nums[idx] > 0:
                nums[idx] *= -1
        return [i + 1 for i, v in enumerate(nums) if v > 0]

Java

class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            int idx = Math.abs(nums[i]) - 1;
            if (nums[idx] > 0) {
                nums[idx] *= -1;
            }
        }
        List<Integer> res = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (nums[i] > 0) {
                res.add(i + 1);
            }
        }
        return res;
    }
}

TypeScript

function findDisappearedNumbers(nums: number[]): number[] {
    for (let i = 0; i < nums.length; i++) {
        let idx = Math.abs(nums[i]) - 1;
        if (nums[idx] > 0) {
            nums[idx] *= -1;
        }
    }
    let ans = [];
    for (let i = 0; i < nums.length; i++) {
        if (nums[i] > 0) {
            ans.push(i + 1);
        }
    }
    return ans;
}

C++

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            int idx = abs(nums[i]) - 1;
            if (nums[idx] > 0)
                nums[idx] *= -1;
        }
        vector<int> res;
        for (int i = 0; i < n; ++i) {
            if (nums[i] > 0)
                res.push_back(i + 1);
        }
        return res;
    }
};

Go

func findDisappearedNumbers(nums []int) []int {
	for _, num := range nums {
		idx := abs(num) - 1
		if nums[idx] > 0 {
			nums[idx] *= -1
		}
	}
	var res []int
	for i, num := range nums {
		if num > 0 {
			res = append(res, i+1)
		}
	}
	return res
}

func abs(a int) int {
	if a > 0 {
		return a
	}
	return -a
}

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