LeetCode #287 FindTheDuplicateNumber. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접, dynamicProgramming
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LeetCode #287
Q.
Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
n+1개의 정수들을 포함하는 정수배열을 입력받는데, 각 정수는 1부터 n 까지이다. 한개의 정수는 반드시 같은게 두개 이상 있다. 딱 한개 중복되는 정수가 있는데 그 수를 찾아라. ( 1~ 4까지 정수면 n+1개, 즉 5개의 정수가 있다. 1개 중복)
노트.
배열 자체를 건드려서는 안된다.
상수만 쓴다? (뭔소리?) 공간복잡도 O(1)
시간복잡도는 O(n^2) 아래여야한다.
딱 한개의 정수만 중복된다. 두개 이상일 수도 있다.
e.g.
Example 1:
Input: [1,3,4,2,2]
Output: 2
Example 2:
Input: [3,1,3,4,2]
Output: 3
Note:
You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.
Process
// Process
//1. Input integers array
//2. Make dynamic map
//3. Fill the dynamic map in
//4. Check the dynamic map over 1
//5. Return index + 1
// 처리과정
//1. 정수배열 입력받는다.
//2. 동적맵 만든다.
//3. 동적맵 채워넣는다.
//4. 동적맵에서 1개 이상인 인덱스 찾는다.
//5. 인덱스 + 1 값 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
#include <vector>
using namespace std;
class Solution {
public:
int findDuplicate(vector<int>& nums) {
bool isDone = false;
int result;
vector<int> dynamic;
for (int i = 0; i < nums.size(); ++i)
dynamic.push_back(0);
for (int i = 0; i < nums.size(); ++i)
++dynamic[nums[i] - 1];
for (int i = 0; !isDone && i < dynamic.size(); ++i)
if (dynamic[i] > 1) {
result = i + 1;
isDone = true;
}
return result;
}
};
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