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.
給出一個(gè)全為正整數(shù)且無(wú)重復(fù)數(shù)字的數(shù)組，找出加起來(lái)是一個(gè)正整數(shù)目標(biāo)值的方法個(gè)數(shù)
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.
這道題可以用dfs+記憶化解決，但是我這里用dp來(lái)解決。思路為：既然target是數(shù)組中數(shù)字的組合，target - nums[i]和target有一定的關(guān)系。假設(shè)dp[i]為數(shù)組的元素加起來(lái)為i的方法個(gè)數(shù)，遞推關(guān)系式為：dp[target] = sum(dp[target - nums[i]])，即把加上每個(gè)nums的元素到達(dá)target的方法數(shù)（dp[target - nums[i]]）相加的總和就是數(shù)組元素加起來(lái)為target的方法總數(shù)
class Solution(object): def combinationSum4(self, nums, target): """ :type nums: List[int] :type target: int :rtype: int """ dp = [1] + [0] * target for index in range(target + 1): # 把每個(gè)target - nums[i]的方法數(shù)相加 for n in nums: # nums的當(dāng)前元素要小于target，否則拋棄該元素 if n <= index: dp[index] += dp[index - n] return dp[target]新聞熱點(diǎn)
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