leetcode 121. Best Time to Buy and Sell Stock

文章目录

• C++
• Java
• Python
• Go

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.

C++

class Solution {
public:
    int maxProfit(vector<int>& a) {
        int minx=INT_MAX,res=0,n=a.size();
        for(int i=0;i<n;i++){
            if(a[i]<minx)
                minx=a[i];
            else if(a[i]-minx>res)
                res=a[i]-minx;
        }
        return res;
    }
};

Java

class Solution {
    public int maxProfit(int[] a) {
        int n=a.length;
        int minx=Integer.MAX_VALUE;
        int res=0;
        for(int i=0;i<n;i++){
            if(a[i]<minx)minx=a[i];
            else if(a[i]-minx>res)res=a[i]-minx;
        }
        return res;
    }
}

Python

class Solution:
    def maxProfit(self, a: List[int]) -> int:
        n,minx,res=len(a),float('inf'),0
        for i in range(n):
            if a[i]<minx:
                minx=a[i]
            elif a[i]-minx>res:
                res=a[i]-minx
        return res

Go

func maxProfit(a []int) int {
    if len(a)<2 {
        return 0
    }
    res,minx:=0,a[0]
    for _,k:=range a {
        if k<minx {
            minx=k
        } else if k-minx>res {
            res=k-minx
        }
    }
    return res
}