448. Find All Numbers Disappeared in an Array

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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.

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[5,6]

思路：数组出现一个数，就在它本应该出现的位置标记（把此位置上的数设置为相反数），如果再出现相同的数，由于已经是负值了，就不再改变。全部标记后，重新扫描数组，未被标记的位置对应一个没有出现过的数（注意：位置 = 数字 - 1）

 1 class Solution {
 2 public:
 3     vector<int> findDisappearedNumbers(vector<int>& nums) {
 4         int len = nums.size();
 5         vector<int> res; //returned list
 6         for (int i = 0; i < len; i++) {
 7             int index = abs(nums[i]) - 1;
 8             if (nums[index] > 0) {
 9                 nums[index] *= -1;
10             }
11         }
12         
13         for (int i = 0; i < len; i++) {
14             if (nums[i] > 0) {
15                 res.push_back(i + 1);
16             }
17         }
18         
19         return res;
20     }
21 };

448. Find All Numbers Disappeared in an Array

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原文地址：http://www.cnblogs.com/xumh/p/7750680.html