题目描述

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

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

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[5,6]

解法一

思路

注意看这题数组元素的值全都在数组索引的范围内,那么肯定就是套用那个经典的给元素值赋正负来表达额外信息的套路了。而这题就是遍历每个元素,把与元素值相等(实际差个1)的那个索引对应的元素赋值为负数,然后遍历一遍完成后,再次遍历,如果某个索引对应的元素值是正数,就说明与这个索引相同的值在数组中不存在,就是结果了。

[4,3,2,7,8,2,3,1]演示一遍过程:

[ 4,  3,  2,  7,  8,  2,  3,  1] -> [ 4,  3,  2, -7,  8,  2,  3,  1]
[ 4,  3,  2, -7,  8,  2,  3,  1] -> [ 4,  3, -2, -7,  8,  2,  3,  1]
[ 4,  3, -2, -7,  8,  2,  3,  1] -> [ 4, -3, -2, -7,  8,  2,  3,  1]
[ 4, -3, -2, -7,  8,  2,  3,  1] -> [ 4, -3, -2, -7,  8,  2, -3,  1]
[ 4, -3, -2, -7,  8,  2, -3,  1] -> [ 4, -3, -2, -7,  8,  2, -3, -1]
[ 4, -3, -2, -7,  8,  2, -3,  1] -> [ 4, -3, -2, -7,  8,  2, -3, -1]
[ 4, -3, -2, -7,  8,  2, -3,  1] -> [ 4, -3, -2, -7,  8,  2, -3, -1]
[ 4, -3, -2, -7,  8,  2, -3,  1] -> [-4, -3, -2, -7,  8,  2, -3, -1]

output: [5, 6]

Python

class Solution(object):
    def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        for num in nums:
            index = num if num > 0 else -num
            index -= 1
            if nums[index] > 0:
                nums[index] = -nums[index]
        result = []
        for (index, num) in enumerate(nums):
            if num > 0:
                result.append(index + 1)
        return result

Java

public class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        for (int num : nums) {
            int index = num > 0 ? num : -num;
            index--;
            if (nums[index] > 0) {
                nums[index] = -nums[index];
            }
        }
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > 0) {
                result.add(i + 1);
            }
        }
        return result;
    }
}

C++

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int> &nums) {
        for (auto num : nums) {
            int index = num > 0 ? num : -num;
            index--;
            if (nums[index] > 0) {
                nums[index] = -nums[index];
            }
        }
        vector<int> result;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] > 0) {
                result.push_back(i + 1);
            }
        }
        return result;
    }
};