448. Find All Numbers Disappeared in an Array
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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]
class Solution { public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> dis = new ArrayList<Integer>(); for(int i = 0;i<nums.length;i++){ int a = Math.abs(nums[i])-1;//利用Math.abs(),避免array中含有相同的数值,多次取负变为正;取a为每个元素对应的数值-1就是为了方便将数值取负,且得到哪些数值没有出现过。 if(nums[a]>0){ nums[a] = -nums[a]; } } for(int i = 0;i<nums.length;i++){ if(nums[i]>0){ dis.add(i+1);//哪一个位置上的元素为正,说明在前述的操作中,没有对应的(位置+1)的元素出现过,否则a=(位置+1)-1,会使得这个位置上的元素变为负的 } } return dis; }}主要思路是假设nums里面的元素用num表示,每一个num的在Array中的位置用location表示。首先遍历nums这个Array,获得每一个num对应的location,然后将num-1位置的数值变为负数。这样,当再次遍历nums的时候,哪一个位置上的数值是正数,就表明哪一个位置+1(即(num-1)+1)这个数是没有出现在nums中的。
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