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]
- public class Solution {
- public List<List<Integer>> combinationSum(int[] candidates, int target) {
- List<List<Integer>> res = new ArrayList<List<Integer>>();
- Arrays.sort(candidates);
- Stack<Integer> stack = new Stack<Integer>();
- dfs(res, stack, target, 0, candidates,0);
- return res;
- }
-
- void dfs(List<List<Integer>> res,Stack<Integer> stack,int target,int cur,int[] arr,int sum){
- if(sum==target){
- ArrayList<Integer> data = new ArrayList<Integer>(stack);
- res.add(data);
- return;
- }else if(sum>target) return;
- else{
- for(int i=cur;i<arr.length;i++){
- stack.add(arr[i]);
- sum += arr[i];
- dfs(res, stack, target, i, arr,sum);
- sum -= arr[i];
- stack.pop();
- }
- }
- }
- }
原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/45725483