题目原址
https://leetcode.com/problems/combination-sum/description/
题目描述
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.
Example1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
解题思路
该题与216题是一个类型题,当然解题思路也是一样。注意:当要求解的结果是一些列集合的集合的时候,要使用DFS搜索记录路径。
AC代码
class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<Integer> list = new ArrayList<Integer>();
List<List<Integer>> ret = new ArrayList<List<Integer>>();
Set<List<Integer>> ret1 = new HashSet<List<Integer>>();
List<List<Integer>> ret2 = new ArrayList<List<Integer>>();
re(candidates, ret, list, 0, 0, target);
return ret;
}
public void re(int[] candidates, List<List<Integer>> ret, List<Integer> list, int sum, int k, int target) {
if(sum == target) {
if(!ret.contains(list))
ret.add(new ArrayList(list));
return;
}
else if(sum > target) return;
for(int i = k; i < candidates.length; i++) {
list.add(candidates[i]);
re(candidates, ret, list, sum + candidates[i], i, target);
list.remove(list.size() - 1);
}
}
}