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]

給出n個數(shù)，其中1～n中有些數(shù)沒出現(xiàn)，求這些沒出現(xiàn)的數(shù)。

我們可以將數(shù)值和數(shù)組的標關(guān)聯(lián)，運用負號巧妙的解決了這個問題。確保每個出現(xiàn)過的數(shù)都變成負數(shù)

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

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轉(zhuǎn)載于:https://www.cnblogs.com/pk28/p/7250740.html

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