题解 | 买卖股票的最好时机(二)(动态规划和贪心算法)

import java.util.Scanner;

// 贪心算法解决问题
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        
        int n = in.nextInt();
        int[] nums = new int[n];
        for(int i = 0; i < n;i++){
            nums[i] = in.nextInt();
        }
        System.out.println(run(n,nums));
    }
     public static int run(int n,int[] nums){
        if(n == 1||n == 0){
            return 0;
        }

         int maxProfit = 0;
        for (int i = 1; i < n; i++) {
            // 只要今天的价格比昨天高，就累加利润
            if (nums[i] > nums[i - 1]) {
                maxProfit += nums[i] - nums[i - 1];
            }
        }

        return maxProfit;
    }
}


/**  // 动态规划二维数组解决问题
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        
        int n = in.nextInt();
        int[] nums = new int[n];
        for(int i = 0; i < n;i++){
            nums[i] = in.nextInt();
        }
        System.out.println(run(n,nums));
    }
     public static int run(int n,int[] nums){
        if(n == 1||n == 0){
            return 0;
        }
        // 在此定义的dp数组的含义为，0为第i天未持有股票，1为第i天持有股票
        int[][] dp = new int[n][2];
        dp[0][0] = 0;
        dp[0][1] = -nums[0];

        for(int i = 1;i < n;i++){
            dp[i][0] = Math.max(dp[i - 1][0] , dp[i - 1][1] + nums[i]);// 未持有股票，就相当于抛出，抛出就要获得当天的钱
            dp[i][1] = Math.max(dp[i - 1][1] , dp[i - 1][0] - nums[i]);// 持有股票，就相当于买入，抛出就要指出当天的钱
        }

        return dp[n-1][0];
    }
}*/