【DP】剑指offer63：股票的最大利润(121.买卖股票的最佳时机)

力扣188（注意：本题与主站 121 题相同：https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/）
动态规划思路：https://leetcode.cn/problems/gu-piao-de-zui-da-li-run-lcof/solutions/192221/mian-shi-ti-63-gu-piao-de-zui-da-li-run-dong-tai-2/

Python

class Solution:
    def bestTiming(self, prices: List[int]) -> int:
        profit = 0
        min_price = inf
        for tmp_price in prices:
            profit = max(profit, tmp_price - min_price)
            min_price = min(min_price, tmp_price)
        
        return profit

Java

class Solution {
    public int bestTiming(int[] prices) {
        int cost = Integer.MAX_VALUE, profit = 0;
        for(int price : prices) {
            cost = Math.min(cost, price);
            profit = Math.max(profit, price - cost);
        }
        return profit;
    }
}

方法1

class Solution {
    public int bestTiming(int[] prices) {
        int size = prices.length;
        if (size < 2) {
            return 0;
        }
        int[] dp = new int[size];         // dp[i]第i天操作可以获取的最大利润
        dp[0] = 0;
        dp[1] = prices[1] - prices[0];
        int max = Math.max(dp[0], dp[1]);
        for (int i = 2; i < size; ++i) {
            int diff = prices[i] - prices[i - 1];
            dp[i] = Math.max(diff + dp[i - 1], diff); // 卖 与 不卖 取较大值
            max = Math.max(max, dp[i]);
        }

        return max;
    }
}

方法2

找最小最大点，好方法！

// 版本1
class Solution {
    public int bestTiming(int[] prices) {
        if (prices.length < 2) {
            return 0;
        }
        int res = 0, minVal = prices[0];
        for (int i = 1; i < prices.length; ++i) {
            minVal = prices[i - 1] < minVal ? prices[i - 1] : minVal;
            res = res > prices[i] - minVal ? res : prices[i] - minVal;
        }

        return res;
    }
}

// 版本2
class Solution {
    
    public int bestTiming(int[] prices) {
        int min = Integer.MAX_VALUE;
        int max = 0;
        for(int i = 0;i<prices.length;i++){
            min = Math.min(prices[i],min);
            max = Math.max(max,prices[i]-min);
        }
        return max;
    }
}