此时我们讲解了0-1背包的多种应用

纯 0 - 1 背包 (opens new window)(动态规划：01背包理论基础（滚动数组）)是求 给定背包容量 装满背包 的最大价值是多少。

package 代码随想录.动态规划.小总结240411;

import java.util.Scanner;

public class _01背包理论基础_滚动数组 {
    public static void main(String[] args) {
        // TODO 假设物品种类为M,背包容量为N
        Scanner scanner = new Scanner(System.in);
        int M = scanner.nextInt();
        int N = scanner.nextInt();
        //对物品赋予其对应该有的价值与重量
        int values [] = new int [M];
        int weights [] = new int [M];
        //给每一个重量与价值赋初始值
        for (int i = 0; i < M;i++){
            weights[i] = scanner.nextInt();
        }
        for (int i = 0;i<M;i++){
            values[i] = scanner.nextInt();
        }
        int dp [][] = new int[M][N+1];
        //初始化dp数组
        for (int i = weights[0];i<=N;i++){
            dp[0][i] = values[0];
        }
        //先遍历物品
        for (int i = 1;i<M;i++){
            //再遍历背包
            for (int j = 0;j<=N;j++){
                if (weights[i] > j){
                    dp[i][j] = dp[i-1][j];  //能背下，则考虑背上去的结果，存储到dp数组中。
                }else {
                    dp[i][j] = Math.max(dp[i-1][j],dp[i-1][j-weights[i]] + values[i]);
                }
            }
        }
        System.out.println(dp[M-1][N]);
    }
}

416. 分割等和子集 (opens new window)是求 给定背包容量，能不能装满这个背包。

package 代码随想录.动态规划.小总结240411;

import java.util.Arrays;

public class _416分割等和子集_dp {
    /**
     *
     * @param nums
     * @return
     */
    public boolean canPartition(int[] nums) {
        // TODO 请你判断是否可以将这个数组分割成两个子集，使得两个子集的元素和相等。
        int len = nums.length;
        if (len == 1){
            return false;
        }
        int allSum = Arrays.stream(nums).sum();
        if (allSum % 2 != 0) {
            return false;
        }
        Arrays.sort(nums);
        int target = allSum / 2;
        //初始化dp数组
        int dp [][] = new int[nums.length][target + 1];
        //初始化dp数组元素
        for(int j = 0; j <= target; j++){
            if(j < nums[0]){
                dp[0][j] = 0;
            }
            else{
                dp[0][j] = nums[0];
            }
        }
//        for(int j = 0; j <= target; j++){
//            dp[0][j] = nums[0];
//        }
        //开始动态规划一下
        for (int i = 1;i<len;i++){
            for (int j  = 0;j<=target;j++){
                if (j < nums[i]){
                    dp[i][j] = dp[i-1][j];
                }else {
                    dp[i][j] = Math.max(dp[i-1][j],dp[i-1][j - nums[i]] + nums[i]);
                }
            }
        }
        //打印出来给你自己看看
        // for(int x : dp){
        //     System.out.print(x + ",");
        // }
        // System.out.print("    "+i+" row"+"\n");
        return dp[len-1][target]  ==  target;
    }

}

1049.最后一块石头的重量 II (opens new window)是求 给定背包容量，尽可能装，最多能装多少.

package 代码随想录.动态规划;

import java.util.Arrays;

public class _1049最后一块石头的重量II_dp {
    /**
     *
     * @param stones
     * @return
     */
    public int lastStoneWeightII(int[] stones) {
        int allSum = 0;
        allSum = Arrays.stream(stones).sum();
        int targetHalf = allSum / 2;
        //初始化，dp[i][j]为可以放0-i物品，背包容量为j的情况下背包中的最大价值
        int [][] dp = new int[stones.length][targetHalf+1];
        //dp[i][0]默认初始化为0
        //dp[0][j]取决于stones[0]
        for (int j = stones[0];j<=targetHalf;j++){
            dp[0][j] = stones[0];
        }

        for (int i = 1;i<stones.length;i++){
            for (int j =1;j<=targetHalf;j++){   //范围为+1，因此可以取到等号
                if (j >= stones[i]){
                    //不放:dp[i - 1][j] 放:dp[i - 1][j - stones[i]] + stones[i]
                    dp[i][j] = Math.max(dp[i-1][j],dp[i-1][j-stones[i]] + stones[i]);
                }else {
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        System.out.println(dp[stones.length - 1][targetHalf]);
        return (allSum - dp[stones.length-1][targetHalf]) - dp[stones.length - 1][targetHalf];
    }
}

 1049. 目标和 (opens new window)是求 给定背包容量，装满背包有多少种方法。

本题是求 给定背包容量，装满背包最多有多少个物品。

package 代码随想录.动态规划;

import java.util.Arrays;

public class _1049最后一块石头的重量II_一维空间压缩 {
    /**
     * 每次都去找最小差
     *
     * @param stones
     * @return
     */
    public int lastStoneWeightII(int[] stones) {
        int sum = 0;
        sum = Arrays.stream(stones).sum();
        int target = sum >> 1;
        //初始化dp数组
        int [] dp = new int[target + 1];
        for (int i = 0; i < stones.length; i++) {
            //采用倒序
            for (int j = target; j >= stones[i];j--){
                //两种情况，要么放，要么不放
                dp[j] = Math.max(dp[j],dp[j - stones[i]] + stones[i]);
            }
        }
        return sum - 2*dp[target];
    }
}

_474一和零_dp

package 代码随想录.动态规划.小总结240411;

public class _474一和零_dp {
    /**
     *
     * @param strs
     * @param m
     * @param n
     * @return
     */
    public int findMaxForm(String[] strs, int m, int n) {
        int [][] dp  = new int[m+1][n+1];
        int length = strs.length;
        for (int i = 0; i < length; i++) {
            int [] zerosAndOnes = getNumsOfZerosAndOnes(strs[i]);
            int zeros = zerosAndOnes[0],ones = zerosAndOnes[1];
            for (int j = m; j >= zeros; j--) {
                for (int k = n; k >= ones; k--){
                    dp[j][k] = Math.max(dp[j][k],dp[j-zeros][k-ones]+1);
                }
            }
        }
        return dp[m][n];
    }

    private int[] getNumsOfZerosAndOnes(String str) {
        int [] zerosOnes = new int[2];
        int length = str.length();
        for (int i = 0;i<length;i++){
            zerosOnes[str.charAt(i) - '0']++;
        }
        return zerosOnes;
    }

}

package 代码随想录.动态规划.小总结240411;

public class _474一和零_三维dp {
    /**
     *
     * @param strs
     * @param m
     * @param n
     * @return
     */
    public int findMaxForm(String[] strs, int m, int n) {
        int length= strs.length;
        int [][][] dp = new int[length+1][m+1][n+1];
        for (int i=1; i<=length; i++){
            int [] zerosAndOnes= getNumsOfZerosAndOnes(strs[i-1]);
            int zeros = zerosAndOnes[0],ones = zerosAndOnes[1];
            for (int j = 0;j<=m;j++){
                for (int k = 0;k<=n;k++){
                    dp[i][j][k] = dp[i-1][j][k];
                    if (j >=zeros &&  k>=ones){
                        dp[i][j][k] = Math.max(dp[i][j][k],dp[i-1][j - zeros][k-ones]+1);
                    }
                }
            }
        }
        return dp[length][m][n];
    }

    private int[] getNumsOfZerosAndOnes(String str) {
        int [] zerosOnes = new  int[2];
        int  length = str.length();
        for (int i = 0;i<length;i++){
            zerosOnes[str.charAt(i)  - '0']++;
        }
        return zerosOnes;
    }

}

_474一和零_三维dp

package 代码随想录.动态规划.小总结240411;

public class _474一和零_三维dp {
    /**
     *
     * @param strs
     * @param m
     * @param n
     * @return
     */
    public int findMaxForm(String[] strs, int m, int n) {
        int length= strs.length;
        int [][][] dp = new int[length+1][m+1][n+1];
        for (int i=1; i<=length; i++){
            int [] zerosAndOnes= getNumsOfZerosAndOnes(strs[i-1]);
            int zeros = zerosAndOnes[0],ones = zerosAndOnes[1];
            for (int j = 0;j<=m;j++){
                for (int k = 0;k<=n;k++){
                    dp[i][j][k] = dp[i-1][j][k];
                    if (j >=zeros &&  k>=ones){
                        dp[i][j][k] = Math.max(dp[i][j][k],dp[i-1][j - zeros][k-ones]+1);
                    }
                }
            }
        }
        return dp[length][m][n];
    }

    private int[] getNumsOfZerosAndOnes(String str) {
        int [] zerosOnes = new  int[2];
        int  length = str.length();
        for (int i = 0;i<length;i++){
            zerosOnes[str.charAt(i)  - '0']++;
        }
        return zerosOnes;
    }

}