442. Find All Duplicates in an Array

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• Total Accepted:?16589
• Total Submissions:?32282
• Difficulty:?Medium
• Contributors:?shen5630

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]

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【題目分析】

給定一個整數數組，數組中的元素都滿足1 =< a[i] <= n。找出數組中重復出現的元素。要求不使用額外的存儲空間，算法的時間復雜度是O(n)。

【思路分析】

對數組中的每個元素，把它當作數組的索引，把該索引對應位置的元素取反，如果發現該位置的元素已經取反，那么這個數就是重復出現的數。

【java代碼】

1 public class Solution { 2 public List<Integer> findDuplicates(int[] nums) { 3 List<Integer> res = new ArrayList<Integer>(); 4 5 for(int i = 0; i < nums.length; i++) { 6 int index = Math.abs(nums[i]) - 1; 7 if(nums[index] < 0) { 8 res.add(index+1); 9 } 10 nums[index] = -nums[index]; 11 } 12 13 return res; 14 } 15 }

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轉載于:https://www.cnblogs.com/liujinhong/p/6416666.html

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