LeetCode 0121. Best Time to Buy and Sell Stock買賣股票的最佳時機【Easy】【Python】【貪心】【動態規劃】
Problem
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.
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.
問題
給定一個數組，它的第 i 個元素是一支給定股票第 i 天的價格。
如果你最多只允許完成一筆交易（即買入和賣出一支股票），設計一個算法來計算你所能獲取的最大利潤。
注意你不能在買入股票前賣出股票。
示例 1:
輸入: [7,1,5,3,6,4]
輸出: 5
解釋: 在第 2 天（股票價格 = 1）的時候買入，在第 5 天（股票價格 = 6）的時候賣出，最大利潤 = 6-1 = 5 。
注意利潤不能是 7-1 = 6, 因為賣出價格需要大于買入價格。
示例 2:
輸入: [7,6,4,3,1]
輸出: 0
解釋: 在這種情況下, 沒有交易完成, 所以最大利潤為 0。
思路
解法一
貪心
買入和賣出同時計算。
當 buy > price，更新買入價格 buy。
當 profit < price - buy，更新利潤 profit。
時間復雜度: O(len(prices))
空間復雜度: O(1)
Python3代碼
class Solution:
def maxProfit(self, prices: List[int]) -> int:
# solution one: 貪心
profit, buy = 0, 0x7FFFFFFF
for price in prices:
if buy > price:
buy = price # 盡量買入價格小的股票
if profit < price - buy:
profit = price - buy # 盡量在最大價格賣出股票
return profit
解法二
動態規劃
找到狀態方程
dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1] + prices[i])
解釋：昨天沒有股票，昨天有股票今天賣出
dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k][0] - prices[i])
解釋：昨天有股票，昨天沒有股票今天買入
base case：
dp[-1][k][0] = dp[i][k][0] = 0
dp[-1][k][1] = dp[i][k][1] = -inf
k = 1
所以 dp[i-1][1][0] = 0
dp[i][1][0] = max(dp[i-1][1][0], dp[i-1][1][1] + prices[i])
dp[i][1][1] = max(dp[i-1][1][1], -prices[i])
k 都是 1，所以 k 對狀態轉移沒有影響：
dp[i][0] = max(dp[i-1][0], dp[i-1][1] + prices[i])
dp[i][1] = max(dp[i-1][1], -prices[i])
i = 0 時，dp[i-1] 不合法。
dp[0][0] = max(dp[-1][0], dp[-1][1] + prices[i])
= max(0, -infinity + prices[i])
= 0
dp[0][1] = max(dp[-1][1], dp[-1][0] - prices[i])
= max(-infinity, 0 - prices[i])
= -prices[i]
空間復雜度: O(1)
Python3代碼
class Solution:
def maxProfit(self, prices: List[int]) -> int:
# solution two: 動態規劃
# K = 1
dp_i_0 = 0
dp_i_1 = float('-inf') # 負無窮
for i in range(len(prices)):
# 昨天沒有股票，昨天有股票今天賣出
dp_i_0 = max(dp_i_0, dp_i_1 + prices[i]) # dp_i_0 和 dp_i_1 可以看成是變量，存儲的都是上一次即昨天的值
# 昨天有股票，昨天沒有股票今天買入
dp_i_1 = max(dp_i_1, -prices[i])
return dp_i_0
