<easy>LeetCode Problem -- 448. Find All Numbers Disappeared in an Array

原创于 2018-02-01 18:26:42 发布 · 167 阅读

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本文介绍了一种在O(n)时间内查找指定范围内未出现在数组中的数字的方法，利用原地哈希的思想，通过修改输入数组来标记已遍历的数字，最终找出所有缺失的数字。

描述：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]

分析：要求找到数组元素为1-n长度为n的数组中为出现的元素。
思路一：采用标记法，存在数字的位置的值置为负数，如果最后值为正数则代表这个数没有出现过。

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        vector<int> res;
        for (int i = 0; i < nums.size(); i++) {
            nums[abs(nums[i]) - 1] = -abs(nums[abs(nums[i]) - 1]);    
            }
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] > 0) res.push_back(i + 1);
        }
        return res;
}
};