Leetcode学习笔记：#448. Find All Numbers Disappeared in an Array

原创于 2019-05-17 20:20:38 发布 · 240 阅读

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该博客围绕Leetcode #448题展开，题目是给定一个整数数组，其中元素范围在1到数组长度n之间，部分元素出现两次，部分出现一次，要找出[1, n]中未在数组里出现的元素，且要求无额外空间、时间复杂度为O(n)，实现思路是首次遍历将见过数字标记为负数，非负数则未遇过并保存。

Leetcode学习笔记：#448. Find All Numbers Disappeared in an Array

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.

实现：

public List<Integer> findDisappearedNumbers(int[] nums){
	List<Integer> ret = new ArrayList<Integer>();

	for(int i = 0; i < nums.length; i++){
		int val = Math.abs(nums[i])-1;
		if(nums[val] > 0){
			nums[val] = -nums[val];
		}
	]
	for(int i = 0; i < nums.length; i++){
		if(nums[i] > 0){
			ret.add(i+1);
		}
	}
	return ret;
}

思路：
第一次遍历，把见过的数字标记为负数，第一次遍历如果遇到不是负数的，说明没有遇过，保存