[LeetCode] 39. Combination Sum_Medium tag: backtracking

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]
]

这个题目思路跟[LeetCode] 40. Combination Sum II_Medium tag: backtracking基础上, 只不过因为没有duplicate,然后sort nums之后,只要nums[i] > target, break 即可。同时因为每个nums[i] 可以用无限次,所以需要在for loop中保持i不变。

Code

class Solution:
    def combinationSum(self, nums, target):
        ans = []
        nums.sort()
        self.search(nums, ans, [], 0, target)
        return ans

    def search(self, nums, ans, temp, pos, target):
        if target == 0:
            ans.append(temp)
        for i in range(pos, len(nums)):
            if nums[i] > target:  # because we sorted the nums
                break
            self.search(nums, ans, temp + [nums[i]], i , target - nums[i]) # keep i instead of i + 1, because we can use one number unlimited times

 

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转载自www.cnblogs.com/Johnsonxiong/p/10924482.html