39. Combination Sum

39. Combination Sum

• 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]]

• 题目大意：给定一个集合，和一个目标数字，找出集合中所有的数，使其和为给出的目标数字。每个数字可以被使用多次。

• 思路：回溯。

• 代码：

  package Array;import java.util.ArrayList;import java.util.List;/** * @Author OovEver * @Date 2017/11/28 10:10 */public class Solution {    public List<List<Integer>> combinationSum(int[] candidates, int target) {        List<List<Integer>> res = new ArrayList<>();        backtrack(res, new ArrayList<>(), candidates, target, 0);        return res;    }    private void backtrack(List<List<Integer>> res, List<Integer> tempList, int[] nums, int target, int start) {        if (target < 0) {            return;        } else if (target == 0) {            res.add(new ArrayList<>(tempList));        } else {            for(int i=start;i<nums.length;i++) {                tempList.add(nums[i]);                backtrack(res, tempList, nums, target - nums[i], i);                tempList.remove(tempList.size() - 1);            }        }    }}