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39. Combination Sum

题目

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

 

Example 1:

Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

Input: candidates = [2], target = 1
Output: []

 

Constraints:

  • 1 <= candidates.length <= 30
  • 2 <= candidates[i] <= 40
  • All elements of candidates are distinct.
  • 1 <= target <= 40
Related Topics
  • 数组
  • 回溯

  • 👍 3046
  • 👎 0
  • 思路

    本题采用穷举来完成,但是可能存在重复元素,限定条件为遍历的起点位置,限制 start 即可

    解法

    py
    # leetcode submit region begin(Prohibit modification and deletion)
    class Solution:
        def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
            res = []
            candidates.sort()
    
            def backtrack(path, total, start):
                if total == target:
                    res.append(path[:])
                    return
                if total > target:
                    return
    
                for i in range(start, len(candidates)):
    
                    path.append(candidates[i])
                    total += candidates[i]
                    backtrack(path, total, i)
                    item = path.pop()
                    total -= item
    
            backtrack([], 0, 0)
    
            return res
            
    # leetcode submit region end(Prohibit modification and deletion)

    复杂度分析

    • 时间复杂度 O(2^n)
    • 空间复杂度 O(target)