Description#
Given an integer array nums of length n where all the integers of nums are in the range [1, n] and each integer appears once or twice, return an array of all the integers that appears twice.
You must write an algorithm that runs in O(n) time and uses only constant extra space.
Example 1:
Input: nums = [4,3,2,7,8,2,3,1]
Output: [2,3]
Example 2:
Input: nums = [1,1,2]
Output: [1]
Example 3:
Input: nums = [1]
Output: []
Constraints:
n == nums.length1 <= n <= 1051 <= nums[i] <= n- Each element in
nums appears once or twice.
Solutions#
Solution 1#
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| class Solution:
def findDuplicates(self, nums: List[int]) -> List[int]:
for i in range(len(nums)):
while nums[i] != nums[nums[i] - 1]:
nums[nums[i] - 1], nums[i] = nums[i], nums[nums[i] - 1]
return [v for i, v in enumerate(nums) if v != i + 1]
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| class Solution {
public List<Integer> findDuplicates(int[] nums) {
int n = nums.length;
for (int i = 0; i < n; ++i) {
while (nums[i] != nums[nums[i] - 1]) {
swap(nums, i, nums[i] - 1);
}
}
List<Integer> ans = new ArrayList<>();
for (int i = 0; i < n; ++i) {
if (nums[i] != i + 1) {
ans.add(nums[i]);
}
}
return ans;
}
void swap(int[] nums, int i, int j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
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| class Solution {
public:
vector<int> findDuplicates(vector<int>& nums) {
int n = nums.size();
for (int i = 0; i < n; ++i) {
while (nums[i] != nums[nums[i] - 1]) {
swap(nums[i], nums[nums[i] - 1]);
}
}
vector<int> ans;
for (int i = 0; i < n; ++i) {
if (nums[i] != i + 1) {
ans.push_back(nums[i]);
}
}
return ans;
}
};
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| func findDuplicates(nums []int) []int {
for i := range nums {
for nums[i] != nums[nums[i]-1] {
nums[i], nums[nums[i]-1] = nums[nums[i]-1], nums[i]
}
}
var ans []int
for i, v := range nums {
if v != i+1 {
ans = append(ans, v)
}
}
return ans
}
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