友情支持
如果您觉得这个笔记对您有所帮助，看在D瓜哥码这么多字的辛苦上，请友情支持一下，D瓜哥感激不尽，😜
|
|
有些打赏的朋友希望可以加个好友，欢迎关注D 瓜哥的微信公众号，这样就可以通过公众号的回复直接给我发信息。
公众号的微信号是: jikerizhi。因为众所周知的原因，有时图片加载不出来。 如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。 |
448. Find All Numbers Disappeared in an Array
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]
参考资料
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] 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
/**
* Runtime: 5 ms, faster than 91.55% of Java online submissions for Find All Numbers Disappeared in an Array.
* Memory Usage: 48.5 MB, less than 33.96% of Java online submissions for Find All Numbers Disappeared in an Array.
*/
public List<Integer> findDisappearedNumbers(int[] nums) {
for (int i = 0; i < nums.length; i++) {
int newIndex = Math.abs(nums[i]) - 1;
if (nums[newIndex] > 0) {
nums[newIndex] *= -1;
}
}
List<Integer> result = new ArrayList<>();
for (int i = 1; i <= nums.length; i++) {
if (nums[i - 1] > 0) {
result.add(i);
}
}
return result;
}
/**
* Runtime: 21 ms, faster than 26.28% of Java online submissions for Find All Numbers Disappeared in an Array.
* Memory Usage: 53.1 MB, less than 15.09% of Java online submissions for Find All Numbers Disappeared in an Array.
*/
public List<Integer> findDisappearedNumbersMap(int[] nums) {
Set<Integer> result = new HashSet<>(nums.length);
for (int i = 1; i <= nums.length; i++) {
result.add(i);
}
for (int num : nums) {
result.remove(num);
}
return new ArrayList<>(result);
}