Star

LeetCode 39. Combination Sum

Posted on | In |

Question

Given a set of candidate numbers (C) (without duplicates) 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. The solution set must not contain duplicate combinations. For example, given candidate set [2, 3, 6, 7] and target 7, A solution set is: 

1
2
3
4
[
  [7],
  [2, 2, 3]
]

Explanation

对于这类求所有组合的题目，必然用DFS来做。注意这里数字是可以重复用的，所以在helper递归的时候还用原来的index。

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> list = new ArrayList<>();
        if (candidates == null || candidates.length == 0) return list;
        Arrays.sort(candidates);
        helper(candidates, 0, target, new ArrayList<Integer>(), list);
        return list;
    }
    // 在list[index:]中选取和为target的所有组合，添加到combination之后，整个放到combinations当中
    public void helper(int[] candidates, int startIndex, int target, List<Integer> combination, List<List<Integer>> list) {
        if (target == 0) list.add(new ArrayList<Integer>(combination));
        for(int i=startIndex; i< candidates.length; i++) {
            if (target < candidates[i]) {
                break;
            }
            combination.add(candidates[i]);
            helper(candidates, i, target - candidates[i], combination,list);
            combination.remove(combination.size()-1);
        }
    }
}