leetcode39. 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]]

解法

回溯法，先对数组进行排序，每次添加本元素及本元素之后的内容。注意ret.add(new ArrayList(temp));要重新new一个list。

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