LeetCode216:Combination Sum III(回溯)

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Note:

  • All numbers will be positive integers.
  • The solution set must not contain duplicate combinations.

Example 1:

Input: k = 3, n = 7
Output: [[1,2,4]]

Example 2:

Input: k = 3, n = 9
Output: [[1,2,6], [1,3,5], [2,3,4]]

LeetCode:链接

216Combination Sum III是在LeetCode40:Combination Sum II(回溯)的基础上,把数组限制为了1-9 9个数字,然后结果的组合也必须是k个数字,和为n。这个的主要不同在于k个数字,其实也就是加个判断。

class Solution(object):
    def combinationSum3(self, k, n):
        """
        :type k: int
        :type n: int
        :rtype: List[List[int]]
        """

        ans, res = [], []
        candidates = list(range(1, 10))
        self.backtracking(candidates, n, 0, ans, res, k)
        return res

    def backtracking(self, candidates, target, start, ans, res, k):
        if target == 0 and len(ans) == k:
            res.append(ans)
        else:
            for i in range(start, len(candidates)):
                if target < candidates[i]:
                    break
                self.backtracking(candidates, target-candidates[i], i+1, ans+[candidates[i]], res, k)

if __name__ == '__main__':
    a = Solution()
    print(a.combinationSum3(2, 18))

猜你喜欢

转载自blog.csdn.net/mengmengdajuanjuan/article/details/84561542