0121. Best Time to Buy and Sell Stock

# 121. Best Time to Buy and Sell Stock#

## 题目 #

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.
``````

## 解题思路 #

• 题目要求找出股票中能赚的钱最多的差价
• 这一题也有多个解法，可以用 DP，也可以用单调栈

## 代码 #

``````
package leetcode

// 解法一 模拟 DP
func maxProfit(prices []int) int {
if len(prices) < 1 {
return 0
}
min, maxProfit := prices, 0
for i := 1; i < len(prices); i++ {
if prices[i]-min > maxProfit {
maxProfit = prices[i] - min
}
if prices[i] < min {
min = prices[i]
}
}
return maxProfit
}

// 解法二 单调栈
func maxProfit1(prices []int) int {
if len(prices) == 0 {
return 0
}
stack, res := []int{prices}, 0
for i := 1; i < len(prices); i++ {
if prices[i] > stack[len(stack)-1] {
stack = append(stack, prices[i])
} else {
index := len(stack) - 1
for ; index >= 0; index-- {
if stack[index] < prices[i] {
break
}
}
stack = stack[:index+1]
stack = append(stack, prices[i])
}
res = max(res, stack[len(stack)-1]-stack)
}
return res
}

`````` Nov 26, 2021 Edit this page