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442.Find-All-Duplicates-in-an-Array

442. Find All Duplicates in an Array

题目地址

题目描述

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
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Find all the elements that appear twice in this array.
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Could you do it without extra space and in O(n) runtime?
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Example:
Input:
[4,3,2,7,8,2,3,1]
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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;
}
}