标签：without style with ++ size numbers ice ber output
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]
正确的序列就藏在下标里。解法很巧妙，利用负号做标记。
class Solution { public: vector<int> findDisappearedNumbers(vector<int>& nums) { vector<int> temp; if (nums.size() == 0) return temp; unsigned int index = 0; for (int i = 0; i < nums.size(); i++) { index = abs(nums[i]) - 1; if (nums[index] > 0) nums[index] = 0 - nums[index]; } for (int i = 0; i < nums.size(); i ++) { if (nums[i] > 0) temp.push_back(i+1); } return temp; } };
Find All Numbers Disappeared in an Array
标签：without style with ++ size numbers ice ber output
原文地址：http://www.cnblogs.com/shawnye/p/6275588.html
