Combination Sum
最后更新于：2022-04-02 01:12:29
# Combination Sum
### Source
- leetcode: [Combination Sum | LeetCode OJ](https://leetcode.com/problems/combination-sum/)
- lintcode: [(135) Combination Sum](http://www.lintcode.com/en/problem/combination-sum/)
~~~
Given a set of candidate numbers (C) and a target number (T),
find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
Have you met this question in a real interview? Yes
Example
given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
Note
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order.
(ie, a1 ≤ a2 ≤ … ≤ ak).
- The solution set must not contain duplicate combinations.
~~~
### 题解
和 [Permutations](http://algorithm.yuanbin.me/zh-cn/exhaustive_search/permutations.html) 十分类似，区别在于剪枝函数不同。这里允许一个元素被多次使用，故递归时传入的索引值不自增，而是由 for 循环改变。
### Java
~~~
public class Solution {
/**
* @param candidates: A list of integers
* @param target:An integer
* @return: A list of lists of integers
*/
public List> combinationSum(int[] candidates, int target) {
List> result = new ArrayList>();
List list = new ArrayList();
if (candidates == null) return result;
Arrays.sort(candidates);
helper(candidates, 0, target, list, result);
return result;
}
private void helper(int[] candidates, int pos, int gap,
List list, List> result) {
if (gap == 0) {
// add new object for result
result.add(new ArrayList(list));
return;
}
for (int i = pos; i < candidates.length; i++) {
// cut invalid candidate
if (gap < candidates[i]) {
return;
}
list.add(candidates[i]);
helper(candidates, i, gap - candidates[i], list, result);
list.remove(list.size() - 1);
}
}
}
~~~
### 源码分析
对数组首先进行排序是必须的，递归函数中本应该传入 target 作为入口参数，这里借用了 Soulmachine 的实现，使用 gap 更容易理解。注意在将临时 list 添加至 result 中时需要 new 一个新的对象。
### 复杂度分析
按状态数进行分析，时间复杂度 O(n!)O(n!)O(n!), 使用了list 保存中间结果，空间复杂度 O(n)O(n)O(n).
### Reference
- Soulmachine 的 leetcode 题解
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