算法--2023.1.25

1.力扣122--买卖股票的最佳时机2

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        int[][] dp = new int[n][2];
        //二维动态规划，
        dp[0][0] = 0;dp[0][1] = -prices[0];
        //dp[i][0]代表当前不持有股票的最大收益，dp[i][1]代表当前持有股票的最大收益
        for(int i = 1;i<n;i++){
            dp[i][0] = Math.max(dp[i-1][0],dp[i-1][1]+prices[i]);
            dp[i][1] = Math.max(dp[i-1][1],dp[i-1][0]-prices[i]);
        }
        return dp[n-1][0];
    }
}

2.力扣309--最佳买卖股票时机含冷冻期

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        int[][] dp = new int[n][3];
        //三维动态规划
        dp[0][0] = 0; dp[0][1] = 0; dp[0][2] = -prices[0];
        for(int i = 1;i<n;i++){
            //dp[i][0]代表卖出的最大值，
            dp[i][0] = Math.max(Math.max(dp[i-1][0],dp[i-1][1]),dp[i-1][2]+prices[i]);
            //dp[i][1]代表冷冻期的值，该值只能是上一轮卖出的值
            dp[i][1] = dp[i-1][0];
            //dp[i][2]代表买入后的最大值，注意实际买入只能在冷冻期后面就是dp[i][1]后面
            dp[i][2] = Math.max(dp[i-1][1]-prices[i],dp[i-1][2]);
        }
        return Math.max(dp[n-1][0],dp[n-1][1]);
    }
}

3.acwing2--01背包问题

　　

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), capcity = in.nextInt();
        int[][] nums = new int[n][2];
        for(int i = 0;i<n;i++){
            int a = in.nextInt(), b = in.nextInt();
            nums[i][0] = a;
            nums[i][1] = b;
        }
        System.out.println(packageProblem(nums,capcity));
    }
    public static int packageProblem(int[][] nums, int capcity){
        int[][] dp = new int[nums.length+1][capcity+1];
        for(int i = 1;i<=nums.length;i++){
            for(int j = 1;j<=capcity;j++){
                
                    dp[i][j] = dp[i-1][j];
                    if(nums[i-1][0]<=j){
                     dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - nums[i - 1][0]] + nums[i - 1][1]);
                    }
            }
        }
        int res = 0;
        for(int i = 1;i<=capcity;i++){
            res = Math.max(res,dp[nums.length][i]);
        }
        return res;
    }

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), capcity = in.nextInt();
        int[][] nums = new int[n][2];
        for(int i = 0;i<n;i++){
            int a = in.nextInt(), b = in.nextInt();
            nums[i][0] = a;
            nums[i][1] = b;
        }
        System.out.println(packageProblem(nums,capcity));
    }
    public static int packageProblem(int[][] nums, int capcity){
        int[] dp = new int[capcity+1];
        for(int i = 1;i<=nums.length;i++){
            for(int j = capcity;j>=nums[i-1][0];j--){
                    dp[j] = Math.max(dp[j-nums[i-1][0]] + nums[i-1][1],dp[j]);
            }
        }
        return dp[capcity];
    }
}

　　

4.acwing3--完全背包问题

import java.util.Scanner;

public class acwing3 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), capcity = in.nextInt();
        int[][] nums = new int[n][2];
        for(int i = 0;i<n;i++){
            nums[i][0] = in.nextInt();
            nums[i][1] = in.nextInt();
        }
        int[] dp = new int[capcity+1];
        for(int i = 0;i<n;i++){
            for(int j = nums[i][0];j<=capcity;j++){
                dp[j] = Math.max(dp[j],dp[j-nums[i][0]] + nums[i][1]);
            }
        }
        System.out.println(dp[capcity]);
    }
}

5.力扣322--零钱兑换

class Solution {
    public int coinChange(int[] coins, int amount) {
    //完全背包问题的动态规划
        int n = coins.length;
        int[] dp = new int[amount+1];
        for(int i = 0;i<=amount;i++){
            dp[i] = 0x3f3f3f3f;
        }
        dp[0] = 0;
        for(int i = 0;i<n;i++){
            //不同轮代表用到的不同硬币
            for(int j = coins[i];j<=amount;j++){
                //dp[j]代表这一轮中容量为j的时候的最小硬币个数
                //因为这里的硬币是可以重复使用的，所以应该从前向后遍历
                dp[j] = Math.min(dp[j],dp[j-coins[i]]+1);
            }
            System.out.println(dp[amount]);
        }
        if(dp[amount] == 0x3f3f3f3f){
            return -1;
        }
        return dp[amount];
    }
}