标签:public ext run numbers could runtime example nts with
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]
数组中的数为1-n,找出缺少的数
C++(238ms):
1 class Solution { 2 public: 3 vector<int> findDisappearedNumbers(vector<int>& nums) { 4 vector<int> res; 5 int len = nums.size() ; 6 for (int i = 0; i < len;i++ ){ 7 int t = abs(nums[i]) - 1 ; 8 if (nums[t] > 0) 9 nums[t] = -nums[t] ; 10 } 11 for (int i = 0; i < len;i++ ){ 12 if (nums[i] >= 0) 13 res.push_back(i+1) ; 14 } 15 return res ; 16 } 17 };
448. Find All Numbers Disappeared in an Array
标签:public ext run numbers could runtime example nts with
原文地址:http://www.cnblogs.com/-Buff-/p/7145264.html