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