40. Combination Sum II

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates may only be used once in the combination.

Note:

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

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

Code：

class Solution(object):
   def combinationSum2(self, candidates, target):
        # Sorting is really helpful, se we can avoid over counting easily
        candidates.sort()                      
        result = []
        self.combine_sum_2(candidates, 0, [], result, target)
        return result
    
   def combine_sum_2(self, nums, start, path, result, target):
        # Base case: if the sum of the path satisfies the target, we will consider 
        # it as a solution, and stop there
        if not target:
            result.append(path)
            return
    
        for i in range(start, len(nums)):
            # Very important here! We don't use `i > 0` because we always want 
            # to count the first element in this recursive step even if it is the same 
            # as one before. To avoid overcounting, we just ignore the duplicates
            # after the first element.
            if i > start and nums[i] == nums[i - 1]:
                continue

            # If the current element is bigger than the assigned target, there is 
            # no need to keep searching, since all the numbers are positive
            if nums[i] > target:
                break

            # We change the start to `i + 1` because one element only could
            # be used once
            self.combine_sum_2(nums, i + 1, path + [nums[i]], 
                            result, target - nums[i])