C,GO,Python3 入門解leetcode 範例122. Best Time to Buy and Sell Stock II

題目出處：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/

給一個趨勢，求利潤。

可以從Example 1, 2, 3 得知 若隔天價格下跌就在今日賣出。

若手邊沒庫存 明日價格又較便宜就不買。若手邊有庫存 明日價格攀升即使有利潤 還是保留不賣。

條件相當多，但是主要還是還是看昨天跟今天或者明天跟今天兩天。

也就是說可以設定兩根指針同時往前走。用這方式來解。

#Ｃ 代碼：

int maxProfit(int* prices, int pricesSize){
    //靜態語言嘛。要先宣告一波。很麻煩。
    //分別是買入價格，總利潤，當日利潤以及庫存
    int buyPrice=0 , profit=0, dayProfit=0;
    bool stock=false;

    //如果只有一天 就不用算了吧
    if (pricesSize <= 1){
        return 0;
    }

    //判斷第一天要不要買入    
    if (prices[0] < prices[1]){
        buyPrice = prices[0];
        stock = true;
    }
    
    //開始遍歷每天去趨勢，兩個指針指向昨天跟今天
    for(int i=1; i<pricesSize; i++){
        if(stock==false && prices[i-1] < prices[i]){
            buyPrice = prices[i-1];
            stock = true;
        }

        //看今天趨勢。有利潤就賣
        if(stock==true && prices[i-1] > prices[i]){
            dayProfit = prices[i-1] - buyPrice;
            profit += dayProfit;
            stock = false;
        }
        
	//如果是最後一天，要執行清倉
        if(i==pricesSize-1 && stock==true){
            dayProfit = prices[i] - buyPrice;
            profit += dayProfit;
            stock = false;
        }
        
    }
    //回傳 總利潤
    return profit;
}

#golang 代碼：

func maxProfit(prices []int) int {
    	var profit, buyPrice int
	var stock bool
	length := len(prices)

	if length == 1 {
		return 0
	}

	if prices[0] < prices[1] {
		buyPrice = prices[0]
		stock = true
	}

	for i := 1; i < length; i++ {

		if stock == false && prices[i-1] < prices[i] {
			buyPrice = prices[i-1]
			stock = true
		}

		if stock == true && prices[i-1] > prices[i] {
			profit += prices[i-1] - buyPrice
			stock = false
		}

		if i == length-1 && stock == true {
			profit += prices[i] - buyPrice
			stock = false
		}

	}

	return profit
}

#Python3 代碼：

class Solution:

    def maxProfit(self, prices: List[int]) -> int:
        profit = 0
        status = False
        days = len(prices)
        
        for i in range(days-1):

            if prices[i]<prices[i+1]:
                if status == False:
                    buy = prices[i]
                    status = True
            else:
                if status:
                    profit += prices[i]- buy
                    status = False


            if i == (days-2) and status:
                profit += prices[i+1] - buy
                return profit
            
        return profit

#Java 給想練的人自己寫，我不寫Java的。

以上。如有問題歡迎提出討論。