LeetCode #540 SingleElementInASortedArray. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
생각나는데로 풀이과정 정리안하고 막풀었더니 코드는 지저분한데 성능은 잘 나옴
LeetCode #540
Q.
Given a sorted array consisting of only integers where every element appears twice except for one element which appears once. Find this single element that appears only once.
정수들만 들어있는 정렬된 배열을 입력 받는데, 배열에는 한개빼고 전부 같은 값이 두개씩 들어있다. 한개만 들어있는 그 값을 찾아봐라.
e.g.
Example 1:
Input: [1,1,2,3,3,4,4,8,8]
Output: 2
Example 2:
Input: [3,3,7,7,10,11,11]
Output: 10
Note: Your solution should run in O(log n) time and O(1) space.
시간복잡도는 O(log n), 공간복잡도는 O(1) 이내로
Process
// Process
//1. Input integer array
//2. Remove specific condition first
// 2.1. Iterate if vector length is enough ( > 3 )
// 2.1.1. If index, index+1, index- value have all different value,
// 2.1.1.1. index value is the integer which have only one element
//3. Return result
// 처리과정
//1. 정수배열을 입력받는다.
//2. 특정 조건값은 미리 제거한다.
// 2.1. 배열 길이가 충분히 길면 반복한다.
// 2.1.1. index, index-1, index+1 값이 전부 다르면
// 2.1.1.1. index 값이 1개인 정수이다.
//3. 구해진 결과 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
#include <vector>
using namespace std;
class Solution {
public:
int singleNonDuplicate(vector<int>& nums) {
int result = 0;
if (nums.size() == 1)
{
result = nums[0];
}
else if (nums.size() == 0)
{
}
else if (nums.size() == 2)
{
if (nums[0] != nums[1])
result = nums[0];
}
else
{
if (nums[0] != nums[1])
{
result = nums[0];
}
else if (nums[nums.size() - 1] != nums[nums.size() - 2])
{
result = nums[nums.size() - 1];
}
else
{
for (int i = 1; i < nums.size() - 2; ++i)
{
if (nums[i] != nums[i - 1] && nums[i] != nums[i + 1])
result = nums[i];
}
}
}
return result;
}
};
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