组合总数

Combination Sum

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。

英文题目:

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 = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

example 2

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

思路

遍历candidates，转化为子问题fun(candidate, target - candidates[i])，如[2, 3, 6, 7], 10，可以先尝试curr = [2]，并转化为子问题[2, 3, 6, 7], 8，递归后为curr = [2, 2]，并再次转化为子问题[2, 3, 6, 7], 6
如遇到和不为target的，回溯

代码

class Solution:
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        result = []
        def combin(candidates, target, i, curr):
            if target == 0:
                result.append(list(curr)) #找到解
                return
            if i >= len(candidates):
                return
            if target < 0:
                return
            if candidates[i] <= target:
                curr.append(candidates[i])                          #尝试一个解
                combin(candidates, target - candidates[i], i, curr) #转化为子问题
                curr.pop()                                          #回溯
                combin(candidates, target, i+1, curr)               #尝试另一个解
            else:
                combin(candidates, target, i+1, curr)
        combin(candidates, target, 0, [])
        return result

本题以及其它leetcode题目代码github地址: github地址