Leetcode 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]]

首先排序是必须的，这没什么好说的
1.想到递归，每一层筛选出一个合理的数，并获得新的target值

2.每一层都要维护一个属于自己的list，传递给下一层的时候创建新的list进行传递(非常重要，每层维护的list容易混淆)

3.后一层都要从前一层选中的数的标号开始往后查找，并且要跳过前后两个数相同的坐标，这样避免出现重复

ps:貌似可以使用深度搜索做.目测效率不高,没有尝试
java代码如下

class Solution {     public List<List<Integer>> combinationSum(int[] candidates, int target) {        Arrays.sort(candidates);        List<List<Integer>> lists = new ArrayList<List<Integer>>();        combineSum(candidates, 0, target, lists, null);        return lists;    }    public void combineSum(int[] candidates, int begin,int target,List<List<Integer>> lists,List<Integer> list) {        for (int i = begin; i < candidates.length; i++) {            if (i > 0 && candidates[i] == candidates[i - 1]) {                continue;            }            int now = candidates[i];            if (now == target) {                if (list == null) {                    list = new ArrayList<Integer>();                }                List<Integer> integerList = new ArrayList<Integer>(list);                integerList.add(now);                lists.add(integerList);                continue;            }            if (now * 2 <=  target) { //两倍的值小于目标值                if (list == null) {                    list = new ArrayList<Integer>();                }                List<Integer> integerList = new ArrayList<Integer>(list);                integerList.add(now);                combineSum(candidates,i,target - now,lists,integerList);            }        }    }}