标签：leetcode and solution length null element put abs i++
Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements of [1, n] inclusive that do not appear in this array.
Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.
Example:
Input: [4,3,2,7,8,2,3,1] Output: [5,6]
public class Solution { public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> arr =new ArrayList<Integer>(); if (nums == null) return arr; for (int i=0; i<nums.length; i++) { int index = abs(nums[i]) - 1; if (nums[index] > 0) nums[index] = -nums[index]; } for (int i=0; i<nums.length; i++) { if (nums[i] > 0) arr.add(i+1); } return arr; } public int abs(int n) { return n < 0 ? -n : n; } }
LeetCode - 448. Find All Numbers Disappeared in an Array
标签：leetcode and solution length null element put abs i++
原文地址：http://www.cnblogs.com/wxisme/p/7307165.html
