LeetCode 41: 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.

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.

For example, given candidate set

2,3,6,7

and
target

7

,

A solution set is:

[7]

[2,
2, 3]

LL's solution:

DFS

public class Solution {
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
public void DFS(int[] candidates,ArrayList<Integer> current,int index,int target){
if(target<0 || index>candidates.length-1)
return;
else if(target==0)
res.add(current);
else{
// left branch, not add the node
DFS(candidates,new ArrayList<Integer>(current),index+1,target);
// right branch, add the node as many as possible
int node = candidates[index];
while(target-node>=0){
current.add(node);
target = target-node;
DFS(candidates,new ArrayList<Integer>(current),index,target);
}
}
return;
}
public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
// Start typing your Java solution below
// DO NOT write main() function
int len = candidates.length;
if(len<1)
return res;
res.clear();
Arrays.sort(candidates);
ArrayList<Integer> current = new ArrayList<Integer>();
DFS(candidates,current,0,target);
HashSet st = new HashSet();
st.addAll(res);
res.clear();
res.addAll(st);
return res;
}
}

Note:

1. DFS做出来res会有重复，因为相同的结果是从不同的路径走出来的，最后要去重（用HashSet）。

2. 要求res里面要从小到大sorted，需保证两点：1. candidate是soted，2. 顺序遍历candidate

若要求从大到小sorted，则只需更改以下几行：

4 if(target<0 || index>candidates.length-1) -> if(target<0 || index<0)

10 DFS(candidates,new ArrayList<Integer>(current),index+1,target); -> index-1

30 DFS(candidates,current,0,target) -> DFS(candidates,current,len-1,target)