[LintCode] Backpack I II III IV V VI [揹包六問]

Backpack I

Problem 單次選擇+最大致積

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

動規經典題目，用數組dp[i]表示書包空間爲i的時候能裝的A物品最大容量。兩次循環，外部遍歷數組A，內部反向遍歷數組dp，若j即揹包容量大於等於物品體積A[i]，則取前i-1次循環求得的最大容量dp[j]，和揹包體積爲j-A[i]時的最大容量dp[j-A[i]]與第i個物品體積A[i]之和即dp[j-A[i]]+A[i]的較大值，做爲本次循環後的最大容量dp[i]。

注意dp[]的空間要給m+1，由於咱們要求的是第m+1個值dp[m]，不然會拋出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 單次選擇+最大價值

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。依然是以揹包空間爲限制條件，所不一樣的是dp[j]取的是價值較大值，而非體積較大值。因此只要把dp[j-A[i]]+A[i]換成dp[j-A[i]]+V[i]就能夠了。

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 重複選擇+最大價值

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 重複選擇+惟一排列+裝滿可能性總數

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 單次選擇+裝滿可能性總數

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 重複選擇+不一樣排列+裝滿可能性總數

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];
    }
}