LeetCode 0039. Combination Sum组合总和【Medium】【Python】【回溯】

Problem

LeetCode

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]
]

问题

力扣

给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的数字可以无限制重复被选取。

说明：

  • 所有数字（包括 target）都是正整数。
  • 解集不能包含重复的组合。

示例 1:

输入: candidates = [2,3,6,7], target = 7,
所求解集为:
[
  [7],
  [2,2,3]
]

示例 2:

输入: candidates = [2,3,5], target = 8,
所求解集为:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

思路

回溯模板

res = []
def backtrack(路径, 选择列表):
    if 满足结束条件:
        res.append(路径)
        return

for 选择 in 选择列表:
    做选择
    backtrack(路径, 选择列表)
    撤销选择

作者：jeromememory
链接：https://leetcode-cn.com/problems/combination-sum/solution/hui-su-suan-fa-tao-mo-ban-ji-ke-by-jeromememory/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
Python3代码
class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        n = len(candidates)
        if n == 0:
            return []

        # accelerate 剪枝提速，非必需
        candidates.sort()

        path, res = [], []
        self.dfs(candidates, 0, n, path, res, target)
        return res

    def dfs(self, candidates, start, n, path, res, target):
        # 1.valid result 递归终止情况
        if target == 0:
            res.append(path[:])
            return

        for i in range(start, n):
            tmp = target - candidates[i]
            # 3.pruning 剪枝
            if tmp < 0:
                break
            # 2.backtrack and update 回溯以及更新 path
            path.append(candidates[i])
            self.dfs(candidates, i, n, path, res, tmp)
            path.pop()

代码地址

GitHub链接

参考

回溯算法 + 剪枝