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文章觀看次數: 747
LeetCode題目翻譯英文原文如下
You are given an array prices where prices[i] is the price of a given stock on the ith day.
You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.
Return the maximum profit you can achieve from this transaction . If you cannot achieve any profit, return 0 .
中文翻譯
給予你一個陣列 prices ,裡面放的是某支股票一段時間的價格 prices[i] = ith 日的價格
你想要透過在一天買進並在未來的某一天賣出,來獲取最高的利潤
回傳你能獲得最高的利潤,如何沒辦法賺到任何收益,則回傳 0
範例及題目限制
範例都比較簡單,總之就是只做一次的買低賣高
範例2 中,因為完全沒有做任何交易故最高收益為 0
解題思路講到股票,就讓人想起了今年,台股一路從萬8 跌了5000多點….
( 大家的損益都綠油油的 )
咳咳,我們先回歸正題 ( 哪天轉型寫金融文章再講講股票 )
看到這一題,大家肯定跟我一樣不到3秒就想出第一個解法了
巢狀迴圈 i 跟 j 給他跑下去 !! 每次確認是否比最大值大這樣
❌超過時間限制
public class Solution {
public int MaxProfit(int[] prices) {
int max = 0;
for(int i = 0;i<prices.Length-1;i++)
{
for(int j = i+1;j<prices.Length;j++)
{
int res = prices[j]-prices[i];
if(res>max)
{
max = res;
}
}
}
return max;
}
}
寫完收工! ➡ 一提交,Error 超過時間限制
時間複雜度 : O(n2 )
這時我們來換一種寫法吧
首先,先設兩個變數 min 跟 max
在迴圈中,要是任一數字比 min 小,就會取代 min
另外保留了比較最大值的設計
相對第一個嘗試,最大的變化就是會動態調整最小值方便計算
整體的時間複雜度也會獲得極大的提升 O(n2 ) ➡ O(n)
C# 解決方案 方案1
public class Solution {
public int MaxProfit(int[] prices) {
int max = 0;
int min = prices[0];
for(int i=1;i<prices.Length;i++){
if(prices[i] < min){
min = prices[i];
}
else if((prices[i] - min) > max )
{
max = prices[i] - min;
}
}
return max;
}
}
時間複雜度 : O(n)
Java解 決方案方案1
class Solution {
public int maxProfit(int[] prices) {
int max = 0;
int min = prices[0];
for(int i=1;i<prices.length;i++){
if(prices[i] < min){
min = prices[i];
}
else if((prices[i] - min) > max )
{
max = prices[i] - min;
}
}
return max;
}
}
Python3 解決方案方案1
class Solution:
def maxProfit(self, prices: List[int]) -> int:
max = 0;
min = prices[0];
for i in range(1,len(prices)):
if prices[i] < min :
min = prices[i]
elif((prices[i] - min) > max):
max = prices[i] - min
return max;
JavaScript 解決方案方案1
/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function(prices) {
var max = 0;
var min = prices[0];
for(let i=1;i<prices.length;i++){
if(prices[i] < min){
min = prices[i];
}
else if((prices[i] - min) > max )
{
max = prices[i] - min;
}
}
return max;
};
結論 要是買股票也有這麼快樂就好了…
這題是 Blind 75 中的一題,算是可以簡單考驗邏輯的題目吧
( Blind 75 之後會額外介紹,簡單來說是一個被大家認為做完就有一定基礎的LeetCode題目清單 )
🧡如果這篇文章有幫上你的一點點忙,那是我的榮幸
🧡可以的話,幫我的FaceBook 粉絲專頁按個讚,我會很感謝的
✅如有任何疑問,歡迎透過留言或messenger讓我知道 !
題目連結 : Best Time to Buy and Sell Stock – LeetCode
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