摘要：單次選擇最大體積動規(guī)經(jīng)典題目，用數(shù)組表示書包空間為的時候能裝的物品最大容量。注意的空間要給，因為我們要求的是第個值，否則會拋出。依然是以背包空間為限制條件，所不同的是取的是價值較大值，而非體積較大值。

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

You can not divide any item into small pieces.

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.

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.

動規(guī)經(jīng)典題目，用數(shù)組dp[i]表示書包空間為i的時候能裝的A物品最大容量。兩次循環(huán)，外部遍歷數(shù)組A，內(nèi)部反向遍歷數(shù)組dp，若j即背包容量大于等于物品體積A[i]，則取前i-1次循環(huán)求得的最大容量dp[j]，和背包體積為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，作為本次循環(huán)后的最大容量dp[i]。

注意dp[]的空間要給m+1，因為我們要求的是第m+1個值dp[m]，否則會拋出OutOfBoundException。

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];
    }
}

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?

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

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.

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

和BackPack I基本一致。依然是以背包空間為限制條件，所不同的是dp[j]取的是價值較大值，而非體積較大值。所以只要把dp[j-A[i]]+A[i]換成dp[j-A[i]]+V[i]就可以了。

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];
    }
}

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?

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

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.

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];
    }
}

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

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

A solution set is:

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

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];
    }
}

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

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

A solution set is:

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

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];
    }
}

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.

The different sequences are counted as different combinations.

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

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];
    }
}