Skip to content

Latest commit

 

History

History
213 lines (169 loc) · 4.84 KB

README_EN.md

File metadata and controls

213 lines (169 loc) · 4.84 KB
comments difficulty edit_url tags
true
Easy
Array
Dynamic Programming

121. Best Time to Buy and Sell Stock

中文文档

Description

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 <= 105
  • 0 <= prices[i] <= 104

Solutions

Solution 1: Enumerate + Maintain the Minimum Value of the Prefix

We can enumerate each element of the array $nums$ as the selling price. Then we need to find a minimum value in front of it as the purchase price to maximize the profit.

Therefore, we use a variable $mi$ to maintain the prefix minimum value of the array $nums$. Then we traverse the array $nums$ and for each element $v$, calculate the difference between it and the minimum value $mi$ in front of it, and update the answer to the maximum of the difference. Then update $mi = min(mi, v)$. Continue to traverse the array $nums$ until the traversal ends.

Finally, return the answer.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

Python3

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        ans, mi = 0, inf
        for v in prices:
            ans = max(ans, v - mi)
            mi = min(mi, v)
        return ans

Java

class Solution {
    public int maxProfit(int[] prices) {
        int ans = 0, mi = prices[0];
        for (int v : prices) {
            ans = Math.max(ans, v - mi);
            mi = Math.min(mi, v);
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int ans = 0, mi = prices[0];
        for (int& v : prices) {
            ans = max(ans, v - mi);
            mi = min(mi, v);
        }
        return ans;
    }
};

Go

func maxProfit(prices []int) (ans int) {
	mi := prices[0]
	for _, v := range prices {
		ans = max(ans, v-mi)
		mi = min(mi, v)
	}
	return
}

TypeScript

function maxProfit(prices: number[]): number {
    let ans = 0;
    let mi = prices[0];
    for (const v of prices) {
        ans = Math.max(ans, v - mi);
        mi = Math.min(mi, v);
    }
    return ans;
}

Rust

impl Solution {
    pub fn max_profit(prices: Vec<i32>) -> i32 {
        let mut ans = 0;
        let mut mi = prices[0];
        for &v in &prices {
            ans = ans.max(v - mi);
            mi = mi.min(v);
        }
        ans
    }
}

JavaScript

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function (prices) {
    let ans = 0;
    let mi = prices[0];
    for (const v of prices) {
        ans = Math.max(ans, v - mi);
        mi = Math.min(mi, v);
    }
    return ans;
};

C#

public class Solution {
    public int MaxProfit(int[] prices) {
        int ans = 0, mi = prices[0];
        foreach (int v in prices) {
            ans = Math.Max(ans, v - mi);
            mi = Math.Min(mi, v);
        }
        return ans;
    }
}

PHP

class Solution {
    /**
     * @param Integer[] $prices
     * @return Integer
     */
    function maxProfit($prices) {
        $ans = 0;
        $mi = $prices[0];
        foreach ($prices as $v) {
            $ans = max($ans, $v - $mi);
            $mi = min($mi, $v);
        }
        return $ans;
    }
}