LeetCode 448. Find All Numbers Disappeared in an Array (在数组中找到没有出现的数字)
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]
Java Solution:
Runtime beats 81.65%
完成日期:05/09/2017
关键词:Array
关键点:把num 和 nums[num - 1] 做1对1的映射
public class Solution
{
public List<Integer> findDisappearedNumbers(int[] nums)
{
List<Integer> dis = new ArrayList<>(); // iterate nums array.
for(int i=0; i<nums.length; i++)
{
int index = Math.abs(nums[i]) - 1; // the nums[i] may be negative.
if(nums[index] > 0) // mark corresponding number to negative if it is positive.
nums[index] *= -1; }
// iterate nums again to add i if nums[i] is positive.
for(int i=0; i<nums.length; i++)
{
if(nums[i] > 0)
dis.add(i+1);
} return dis;
}
}
参考资料:
www.cnblogs.com/grandyang/p/6222149.html
LeetCode 题目列表 - LeetCode Questions List
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