Middle-题目54：39. Combination Sum

题目原文：
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]
题目大意：
给出一系列候选数，及一个目标数，计算目标数拆分成候选数之和（允许重复）的所有拆分方式。
题目分析：
贪心+回溯。使用backtrack(List<List<Integer>> list, List<Integer> sublist, int[] candidates, int target, int startPos)维护搜索过程，参数意义不必过多介绍，每次加入一个候选数，就从当前候选数开始往下搜。
源码：（language：java）

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> list = new ArrayList<List<Integer>>();
        backtrack(list, new ArrayList<Integer>(), candidates, target, 0);
        return list;
    }
    private void backtrack(List<List<Integer>> list, List<Integer> sublist, int[] candidates, int target, int startPos) {
        if(target == 0) {
            list.add(new ArrayList<Integer>(sublist));
        }
        else {
            for(int i = startPos; i < candidates.length; i++) {
                int temp = candidates[i];
                if(temp <= target) {
                    sublist.add(temp);
                    backtrack(list,sublist,candidates,target-temp,i);
                    sublist.remove(sublist.size()-1);
                }
            }
        }
    }
}

成绩：
6ms,beats 64.44%,众数5ms,24.50%