LeetCode 39

LeetCode 39. Combination Sum

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

解题思路

该程序需要求出给定数组中数字的组合，使得能够正好组成给定的和（target），并且数组中的数字能够重复使用。解决该问题能够用到递归的方法，即首先对数组由大到小排序，遍历数组中的每一个数，每一层遍历生成更接近于target的数的组合。在每一层递归中，若组合超过了target，则忽略该组合（不符合要求）。若组合正好为target，则添加该组合进入答案（ans）。若组合小于target，则继续递归到下一层。最终答案存储在列表ans中。

源代码

class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        candidates.sort()
        ans = []
        self.dfs(candidates, target, 0, [], ans)
        return ans

    def dfs(self, arrays, target, pos, path, ans):
        if target < 0:
            return
        if target == 0:
            ans.append(path)
            return
        for i in range(pos, len(arrays)):
            self.dfs(arrays, target - arrays[i], i, path + [arrays[i]], ans)