# LeetCode Solution, Easy, 121. Best Time to Buy and Sell Stock

# [121. Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)

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`.

**Example 1:**

    Input: prices = [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.
    Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

**Example 2:**

    Input: prices = [7,6,4,3,1]
    Output: 0
    Explanation: In this case, no transactions are done and the max profit = 0.

**Constraints:**

-   `1 <= prices.length <= 10^5`
-   `0 <= prices[i] <= 10^4`

### 題目敘述

#### 題目翻譯

題目就是會給一個陣列 `prices`，其中的每個元素都表示當天股票的價格。然後要找出在一次買賣的操作中可以獲得的最大獲利。如果沒有辦法獲利的話就回傳 `0`。

### 解法解析

這題很簡單的解法就是直接找出這個陣列的最大值和最小值，相減就會是最大的獲利了。唯一需要注意的是因為買賣操作，一定是買先後賣。所以順序的限制要注意到。

#### 程式範例

##### Go

```go
func maxProfit(prices []int) int {
	minPrice := math.MaxInt32
	theMaxProfit := 0
	for _, price := range prices {
		if price < minPrice {
			minPrice = price
		} else if price-minPrice > theMaxProfit {
			theMaxProfit = price - minPrice
		}
	}

	return theMaxProfit
}
```

##### JavaScript

```javascript
/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function (prices) {
    let minPrice = Number.MAX_SAFE_INTEGER,
        theMaxProfit = 0;
    for (let price of prices) {
        if (price < minPrice) {
            minPrice = price;
        } else if (price - minPrice > theMaxProfit) {
            theMaxProfit = price - minPrice;
        }
    }

    return theMaxProfit;
};
```

##### Kotlin

```kotlin
class Solution {
    fun maxProfit(prices: IntArray): Int {
        var minPrice = Int.MAX_VALUE
        var theMaxProfit = 0

        for (price in prices) {
            if (price < minPrice) {
                minPrice = price
            } else if (price - minPrice > theMaxProfit) {
                theMaxProfit = price - minPrice
            }
        }

        return theMaxProfit
    }
}
```

##### PHP

```php
class Solution
{

    /**
     * @param Integer[] $prices
     * @return Integer
     */
    function maxProfit($prices)
    {
        $minPrice = PHP_INT_MAX;
        $maxProfit = 0;

        foreach ($prices as $price) {
            if ($price < $minPrice) {
                $minPrice = $price;
            } elseif ($price - $minPrice > $maxProfit) {
                $maxProfit = $price - $minPrice;
            }
        }

        return $maxProfit;
    }
}
```

##### Python

```python
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        min_price = float('inf')
        max_profit = 0
        for i in range(len(prices)):
            if prices[i] < min_price:
                min_price = prices[i]
            elif prices[i] - min_price > max_profit:
                max_profit = prices[i] - min_price

        return max_profit
```

##### Rust

```rust
impl Solution {
    pub fn max_profit(prices: Vec<i32>) -> i32 {
        let mut max_profit = 0;
        let mut min_price = std::i32::MAX;
        for price in prices {
            if price < min_price {
                min_price = price;
            } else if price - min_price > max_profit {
                max_profit = price - min_price;
            }
        }
        max_profit
    }
}
```

##### Swift

```swift
class Solution {
    func maxProfit(_ prices: [Int]) -> Int {
        var minPrice = Int.max
        var theMaxProfit = 0

        for price in prices {
            if price < minPrice {
                minPrice = price
            } else if price - minPrice > theMaxProfit {
                theMaxProfit = price - minPrice
            }
        }

        return theMaxProfit
    }
}
```