Backpack VI

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

Example:

nums = [1, 2, 3]
target = 4

The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)

Note that different sequences are counted as different combinations.

Therefore the output is 7.

答案
Backpack题是要找出最大不超过target的物品重量sum
Backpack V题是要找出有多少种物品重量sum刚好等于target
而这个Backpack VI题是要找出有多少种物品重量sum刚好等于target，每个物品可以无限用，而且只要用的顺序不一样就算一种组合

dfs+memorization

class Solution {
    int[] dp = null;
    public int combinationSum4(int[] nums, int target) {
        if(target == 0) return 1;
        
        int ans = 0;
        dp = new int[target + 1];
        Arrays.fill(dp, -1);
        dp[0] = 1;
        recur(nums, target);
        return dp[target];
    }
    
    public int recur(int[] nums, int target) {
        // if(target == 0) return 1; This line is not needed if you have initilized dp[0] = 1
        if(dp[target] != -1) return dp[target];
        
        int ans = 0;
        for(int i = 0; i < nums.length; i++) {
            if(target >= nums[i]) {
                ans += recur(nums, target - nums[i]);                
            }
        }
        
        dp[target] = ans;
        return ans;
    }
}

dp

public class Solution {
    /**
     * @param nums: an integer array and all positive numbers, no duplicates
     * @param target: An integer
     * @return: An integer
     */
    public int backPackVI(int[] nums, int target) {
        int n = nums.length, m = target;
        if(n == 0) return 0;
        
        int[] f = new int[target + 1];
        f[0] = 1;

        for(int i = 1; i <= m; i++) {
            for(int j = 0; j < n; j++) {
                if(i - nums[j] >= 0)
                    f[i] = f[i] + f[i - nums[j]];
            }
        }
        return f[m];
    }
}