Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements that appear twice in this array.
Could you do it without extra space and in O(n) runtime?
Example:
Input:
[4,3,2,7,8,2,3,1]
Output:
[2,3]
代码
Approach #1 Turn each number negative
class Solution {
public List<Integer> findDuplicates(int[] nums) {
List<Integer> res = new ArrayList();
for (int i = 0; i < nums.length; i++) {
int index = Math.abs(nums[i]) - 1;
if (nums[index] < 0) {
res.add(Math.abs(index + 1));
}
nums[index] = -nums[index];
}
return res;
}
}
Approach #2 Find All Numbers Disappeared in an Array
class Solution {
public List<Integer> findDisappearedNumbers(int[] nums) {
List<Integer> res = new ArrayList();
for (int num : nums){
int index = Math.abs(num) - 1;
if (nums[index] > 0) nums[index] = -nums[index];
}
for (int i = 0; i < nums.length; i++){
if (nums[i] > 0) res.add(i + 1);
}
return res;
}
}