LeetCode #0448 FindAllNumbersDisappearedInAnArray. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
#442 랑 거의 비슷함, 성능은 그닥..
It's almost similar with #442, poor performance..
Runtime: 116 ms, faster than 66.91% of C++ online submissions for Find All Numbers Disappeared in an Array.
Memory Usage: 17.6 MB, less than 6.67% of C++ online submissions for Find All Numbers Disappeared in an Array.
LeetCode #448
Q.
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.
1 <= a[i] <= n (n은 배열길이) 사이값이 들어가있는 정수 배열에서, 몇몇 값들은 두번 나오고 어떤것들은 한번 나온다.
1~ n 까지의 정수배열에서 배열 안에 한번도 나오지 않은 값을 찾아라
메모리 쓰지말고, O(n) 시간복잡도 가능? 리턴되는 list는 extra로 안침.
Example:
Input: [4,3,2,7,8,2,3,1]
Output: [5,6]
Process
// Process
//1. Input array
//2. Make dynamicArray
//3. Iterate from begin to the end of dynamicArray
// 3.1. Count ++ index [inputArray[i]]
//4. Iterate from begin to the end of dynamicArray
// 4.1. Check if it's 0
// 4.1.1. If so -> Add to resultArray
//5. Return resultArray
// 처리과정
//1. 배열 입력받는다
//2. 다이나믹배열 만들어둔다.
//3. 만든 다이나믹배열 반복한다.
// 3.1. 다이나믹 배열의 inputArray[i] 번째 인덱스에 1씩 더한다.
//4. 다이나믹 배열 반복한다.
// 4.1. 0인지 확인해서
// 4.1.1. 0이면 resultArray 에 더한다.
//5. resultArray 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
vector<int> findDisappearedNumbers(vector<int>& nums) {
vector<int> resultVec;
const int length = nums.size();
vector<int> dynamicBuffer;
for (int i = 0; i < length; ++i) {
dynamicBuffer.push_back(0);
}
for (int i = 0; i < length; ++i) {
++dynamicBuffer[nums[i]-1];
}
for (int i = 0; i < length; ++i) {
if (dynamicBuffer[i] == 0) {
resultVec.push_back(i + 1);
}
}
return resultVec;
}
};
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