【LeetCode】442. Find All Duplicates in an Array

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问题描述

问题链接：https://leetcode.com/problems/binary-watch/#/description

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]

我的代码

思路的话之前好像看过类似的，就用了。核心就是在原来的数组里面藏一些信息。

public class Solution {    public List<Integer> findDuplicates(int[] nums) {        /*        思路：第1趟的时候为nums[nums[i]]加上一个n，然后选用的时候用num = nums[i] % n == 0?n:nums[i] % n来取数        第2趟的时候凡是>2n的就是出现2次的。        */        List<Integer> result = new ArrayList<Integer>();        int len = nums.length;        int n = len;        for(int i = 0; i < len; i++){            int num = nums[i] % n == 0?n:nums[i] % n;            nums[num - 1] += len;        }        for(int i = 0; i < len; i++){            if(nums[i] > 2 * n){                result.add(i + 1);            }        }        return result;    }}

打败了58.73%的Java代码。来看看讨论区。

讨论区

Java Simple Solution

链接地址：https://discuss.leetcode.com/topic/64735/java-simple-solution

基本是我那个思路。

public class Solution {    // when find a number i, flip the number at position i-1 to negative.     // if the number at position i-1 is already negative, i is the number that occurs twice.    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;    }}

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