問題:Givenan array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appearonce.Find all the elements of [1, n] inclusive that do not appear in thisarray.Could you do it without extra space and in O(n) runtime? You may assumethe returned list does not count as extra space.
Example: Input:[4,3,2,7,8,2,3,1] Output:[5,6]
思考:現将數組按元素值,從小到大排序,并建立一個vectorresult儲存最後的輸出。如果第一個元素非1,則将從1到nums[0]-1的值都存在result數組裡面。接着看中間的元素,如果某一個值與前一個值相等,或者等于前一個值加1,則可繼續。若大于,則将這其中少的幾個值存在result數組中。再看最後一個值nums[n-1],将nums[n-1]和n之間值存在result中。
代碼:
class Solution {
public:
vector<int> findDisappearedNumbers(vector<int>& nums) {
sort(nums.begin(),nums.end());
int n=nums.size();
vector<int> result;
if(!n) return result;
int idx=1;
//1和nums[0]之間的
while(nums[0]>idx)
result.push_back(idx++);
//中間部分
for(int i=1;i<n;i++)
{
if(nums[i]==nums[i-1]||nums[i]==nums[i-1]+1)
continue;
if(nums[i]>nums[i-1]+1)
{
for(int j=nums[i-1]+1;j<nums[i];j++)
result.push_back(j);
}
}
idx=nums[n-1];
//nums[n-1]和n之間
while(idx!=n)
{
result.push_back(++idx);
}
return result;
}
};