LeetCode #121 BestTimeToBuyAndSellStock. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
퍼포먼스 최악
The worst performance
LeetCode #121
Q.
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
i 날짜마다 물건의 가격이 정해진 배열이 있다.
만약 한번에 거래만 허용된다면, (예를들면, 한번 사고 한번 팔 수 있다.)
최대 수익을 낼 수 있는 알고리즘을 짜봐라.
사기 전에는 팔 수 없다.
e.g.
Example 1:
Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. Not 7-1 = 6, as selling price needs to be larger than buying price.
Example 2:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
Process
// Process
//1. Input prices vector
//2. Iterate from begin to the end - 1
// 2.1. Iterate from i+1 to the end
// 2.1.1. Compare profit with maxProfit
// 2.1.1.1. Replace maxProfit if profit is bigger than maxProfit
//3. Return maxProfit
// 처리과정
//1. 가격들이 있는 벡터를 입력받는다.
//2. 시작부터 끝-1 까지 반복한다.
// 2.1. i+1 부터 끝까지 반복한다.
// 2.1.1. profit 과 maxProfit 비교해서
// 2.1.1.1. profit이 더 크면 profit을 maxProfit 으로 치환한다.
//3. maxProfit 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
int maxProfit(vector<int>& prices) {
int maxProfit = 0;
int length = prices.size();
if (length > 0) {
for (int i = 0; i < length - 1; ++i) {
for (int j = i + 1; j < length; ++j) {
if (prices[j] - prices[i] > maxProfit) {
maxProfit = prices[j] - prices[i];
}
}
}
}
return maxProfit;
}
};
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