leetcode Combination Sum IV

leetcode Combination Sum IV

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:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
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.
Follow up: What if negative numbers are allowed in the given array? How does it change the problem? What limitation we need to add to the question to allow negative numbers?

题目地址：leetcode Combination Sum IV

题意：给定一个元素互不相同且均为正数的数组，让你用该数组中的数组合成target（可以重复使用），问有多少种。

思路：DP可以有两种

dp[i] += dp[i-num]
dp[i+num] += dp[i]

其实和 322 Coin Change 那题差不多。。。

第一种DP

C++

class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        vector<int> dp(target+1,0);
        dp[0] = 1;
        for(int i = 1;i <= target ;i++){
            for(int num:nums){
                if(i >= num) dp[i] += dp[i-num];
            }
        }
        return dp[target];
    }
};

Java

public class Solution {
    public int combinationSum4(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(i >= num) dp[i] += dp[i - num];
            }
        }
        return dp[target];
    }
}

Python

class Solution(object):
    def combinationSum4(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        dp = [1] + [0] * target
        for i in xrange(1, target + 1):
            for x in nums:
                if i >= x:
                    dp[i] += dp[i - x]
        return dp[target]

第二种DP C++

class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        vector<int> dp(target+1,0);
        dp[0] = 1;
        for(int i = 0; i < target;i++){
            for(int num:nums){
                if(i + num <= target) dp[i + num] += dp[i];
            }
        }
        return dp[target];
    }
};

Java

public class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp= new int[target+1];
        dp[0] = 1;
        for(int i = 0; i < target;i++){
            for(int num:nums){
                if(i + num <= target) dp[i + num] += dp[i];
            }
        }
        return dp[target];
    }
}

Python

class Solution(object):
    def combinationSum4(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        dp = [1] + [0] * target
        for i in xrange(target + 1):
            for x in nums:
                if i + x <= target:
                    dp[i + x] += dp[i]
        return dp[target]

本题是leetcode 377 Combination Sum IV题解

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