[LintCode] Backpack I II III IV V VI [背包六问]

Backpack I

Problem 单次选择+最大致积

Given n items with size Ai, an integer m denotes the size of a backpack. How full you can fill this backpack?

Notice

You can not divide any item into small pieces.

Example

If we have 4 items with size [2, 3, 5, 7], the backpack size is 11, we can select [2, 3, 5], so that the max size we can fill this backpack is 10. If the backpack size is 12. we can select [2, 3, 7] so that we can fulfill the backpack.

You function should return the max size we can fill in the given backpack.

Challenge

O(n x m) time and O(m) memory.

O(n x m) memory is also acceptable if you do not know how to optimize memory.

Note

动规经典题目，用数组dp[i]表示书包空间为i的时候能装的A物品最大容量。两次循环，外部遍历数组A，内部反向遍历数组dp，若j即背包容量大于等于物品体积A[i]，则取前i-1次循环求得的最大容量dp[j]，和背包体积为j-A[i]时的最大容量dp[j-A[i]]与第i个物品体积A[i]之和即dp[j-A[i]]+A[i]的较大值，做为本次循环后的最大容量dp[i]。

注意dp[]的空间要给m+1，由于咱们要求的是第m+1个值dp[m]，不然会抛出OutOfBoundException。

Solution

public class Solution {
    public int backPack(int m, int[] A) {
        int[] dp = new int[m+1];
        for (int i = 0; i < A.length; i++) {
            for (int j = m; j > 0; j--) {
                if (j >= A[i]) {
                    dp[j] = Math.max(dp[j], dp[j-A[i]] + A[i]);
                }
            }
        }
        return dp[m];
    }
}

Backpack II

Problem 单次选择+最大价值

Given n items with size A[i] and value V[i], and a backpack with size m. What's the maximum value can you put into the backpack?

Notice

You cannot divide item into small pieces and the total size of items you choose should smaller or equal to m.

Example

Given 4 items with size [2, 3, 5, 7] and value [1, 5, 2, 4], and a backpack with size 10. The maximum value is 9.

Challenge

O(n x m) memory is acceptable, can you do it in O(m) memory?

Note

和BackPack I基本一致。依然是以背包空间为限制条件，所不一样的是dp[j]取的是价值较大值，而非体积较大值。因此只要把dp[j-A[i]]+A[i]换成dp[j-A[i]]+V[i]就能够了。

Solution

public class Solution {
    public int backPackII(int m, int[] A, int V[]) {
        int[] dp = new int[m+1];
        for (int i = 0; i < A.length; i++) {
            for (int j = m; j > 0; j--) {
                if (j >= A[i]) dp[j] = Math.max(dp[j], dp[j-A[i]]+V[i]);
            }
        }
        return dp[m];
    }
}

Backpack III

Problem 重复选择+最大价值

Given n kind of items with size Ai and value Vi( each item has an infinite number available) and a backpack with size m. What's the maximum value can you put into the backpack?

Notice

You cannot divide item into small pieces and the total size of items you choose should smaller or equal to m.

Example

Given 4 items with size [2, 3, 5, 7] and value [1, 5, 2, 4], and a backpack with size 10. The maximum value is 15.

Solution

public class Solution {
    public int backPackIII(int[] A, int[] V, int m) {
        int[] dp = new int[m+1];
        for (int i = 0; i < A.length; i++) {
            for (int j = 1; j <= m; j++) {
                if (j >= A[i]) dp[j] = Math.max(dp[j], dp[j-A[i]]+V[i]);
            }
        }
        return dp[m];
    }
}

Backpack IV

Problem 重复选择+惟一排列+装满可能性总数

Given n items with size nums[i] which an integer array and all positive numbers, no duplicates. An integer target denotes the size of a backpack. Find the number of possible fill the backpack.

Each item may be chosen unlimited number of times

Example

Given candidate items [2,3,6,7] and target 7,

A solution set is:

[7]
[2, 2, 3]
return 2

Solution

public class Solution {
    public int backPackIV(int[] nums, int target) {
        int[] dp = new int[target+1];
        dp[0] = 1;
        for (int i = 0; i < nums.length; i++) {
            for (int j = 1; j <= target; j++) {
                if (nums[i] == j) dp[j]++;
                else if (nums[i] < j) dp[j] += dp[j-nums[i]];
            }
        }
        return dp[target];
    }
}

Backpack V

Problem 单次选择+装满可能性总数

Given n items with size nums[i] which an integer array and all positive numbers. An integer target denotes the size of a backpack. Find the number of possible fill the backpack.

Each item may only be used once

Example

Given candidate items [1,2,3,3,7] and target 7,

A solution set is:

[7]
[1, 3, 3]
return 2

Solution

public class Solution {
    public int backPackV(int[] nums, int target) {
        int[] dp = new int[target+1];
        dp[0] = 1;
        for (int i = 0; i < nums.length; i++) {
            for (int j = target; j >= 0; j--) {
                if (j >= nums[i]) dp[j] += dp[j-nums[i]];
            }
        }
        return dp[target];
    }
}

Backpack VI aka: Combination Sum IV

Problem 重复选择+不一样排列+装满可能性总数

Given an integer array nums with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Notice

The different sequences are counted as different combinations.

Example

Given nums = [1, 2, 4], target = 4

The possible combination ways are:

[1, 1, 1, 1]
[1, 1, 2]
[1, 2, 1]
[2, 1, 1]
[2, 2]
[4]
return 6

Solution

public class Solution {
    public int backPackVI(int[] nums, int target) {
        int[] dp = new int[target+1];
        dp[0] = 1;
        for (int i = 1; i <= target; i++) {
            for (int num: nums) {
                if (num <= i) dp[i] += dp[i-num];
            }
        }
        return dp[target];
    }
}