## 题目描述

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.
``````

## 关键点解析

• 这类题只要你在心中（或者别的地方）画出上面这种图就很容易解决

## 代码

C++ 解法：

``````class Solution {
public:
int maxProfit(vector<int>& prices) {
int res = 0, buy = INT_MAX;
for (int price : prices) {
res = max(res, price - buy);
}
return res;
}
};
``````

Java 解法：

``````public class Solution {
public int maxProfit(int[] prices) {
int res = 0, buy = Integer.MAX_VALUE;
for (int price : prices) {
res = Math.max(res, price - buy);
}
return res;
}
}
``````

### 思路2

JS Code:

``````/*
* @lc app=leetcode id=121 lang=javascript
*
* [121] Best Time to Buy and Sell Stock
*
*
* algorithms
* Easy (46.34%)
* Total Accepted:    480.5K
* Total Submissions: 1M
* Testcase Example:  '[7,1,5,3,6,4]'
*
* Say you have an array for which the i^th 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.
*
*
*/
/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function(prices) {
let min = prices[0];
let profit = 0;
// 7 1 5 3 6 4
for(let i = 1; i < prices.length; i++) {
if (prices[i] > prices[i -1]) {
profit = Math.max(profit, prices[i] - min);
} else {
min = Math.min(min, prices[i]);;
}
}

return profit;
};
``````

Python Code:

``````class Solution:
def maxProfit(self, prices: 'List[int]') -> int:
if not prices: return 0

min_price = float('inf')
max_profit = 0

for price in prices:
if price < min_price:
min_price = price
elif max_profit < price - min_price:
max_profit = price - min_price
return max_profit
``````

C++ Code:

``````/**
* 系统上C++的测试用例中的输入有[]，因此需要加一个判断
*/
class Solution {
public:
int maxProfit(vector<int>& prices) {
if (prices.empty()) return 0;
auto min = prices[0];
auto profit = 0;
for (auto i = 1; i < prices.size(); ++i) {
if (prices[i] > prices[i -1]) {
profit = max(profit, prices[i] - min);
} else {
min = std::min(min, prices[i]);;
}
}
return profit;
}
};
``````

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