leetcode-448 Find All Numbers Disappeared in an Array
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题意:给一个长度为n,1《元素值《n,求数值1~n中没出现过的数,时间复杂度限制为O(n),除了要输出的数组,不能开额外的空间
思路:由于时间和空间的限制,所以只能在原来的数组上进行操作,由于元素值为1~n,正好对应数组的下标,所以对每个出现过的数值,把其对应的下标元素变为负值,最后遍历一遍数组,元素值大于0的下标就是没出现过的数。
class Solution {public: vector<int> findDisappearedNumbers(vector<int>& nums) { vector<int> arr; 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) { arr.push_back(i+1); } } return arr; }};
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