LeetCode 121. Best Time to Buy and Sell Stock--Java,Python,C++解法

LeetCode 121. Best Time to Buy and Sell Stock
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タイトルアドレス:Best Time to Buy and Sell Stock-LeetCode
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.

この問題は一度遍歴すればよい.Pythonの解法は以下の通りです.

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        maxprofit=0
        minprice=-1
        for i in prices:
            if minprice==-1:
                minprice=i
            elif i<minprice:
                minprice=i
            elif i-minprice>maxprofit:
                maxprofit=i-minprice
        return maxprofit

Javaの解法は次のとおりです.

public class Solution {
    public int maxProfit(int prices[]) {
        int minprice = Integer.MAX_VALUE;
        int maxprofit = 0;
        for (int i = 0; i < prices.length; i++) {
            if (prices[i] < minprice)
                minprice = prices[i];
            else if (prices[i] - minprice > maxprofit)
                maxprofit = prices[i] - minprice;
        }
        return maxprofit;
    }
}

C++解法は以下の通りである.

class Solution {
public:
int maxProfit(vector<int> &prices) {
    int maxPro = 0;
    int minPrice = INT_MAX;
    for(int i = 0; i < prices.size(); i++){
        minPrice = min(minPrice, prices[i]);
        maxPro = max(maxPro, prices[i] - minPrice);
    }
    return maxPro;
}
};