[leetcode]39. Combination Sum(Java)

leetcode：https://leetcode.com/problems/combination-sum/#/description

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: 

[  [7],  [2, 2, 3]]

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JavaCode：

package go.jacob.day621;import java.util.ArrayList;import java.util.Arrays;import java.util.List;/** * [leetcode]39. Combination Sum *  * @author Administrator * */public class Demo3 {public List<List<Integer>> combinationSum(int[] candidates, int target) {List<List<Integer>> res = new ArrayList<List<Integer>>();if (candidates == null || candidates.length < 1)return res;Arrays.sort(candidates);backtrack(candidates, target, res, new ArrayList<Integer>(), 0);return res;}// 如果不加参数start，会出现：input:[2,3,6,7]7// 结果：[[2,2,3],[2,3,2],[3,2,2],[7]]instead of [[2,2,3],[7]]的情况private void backtrack(int[] nums, int target, List<List<Integer>> res, ArrayList<Integer> list, int start) {if (target < 0)return;if (target == 0) {res.add(list);return;}//加上target >= nums[i]：33ms to 23msfor (int i = start; i < nums.length && target >= nums[i]; i++) {list.add(nums[i]);backtrack(nums, target - nums[i], res, new ArrayList<Integer>(list), i);list.remove(list.size() - 1);}}}