Combination Sum II

Combination Sum II

问题：

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

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.

思路：

　　常见的回溯问题

我的代码：

public class Solution {

    public List<List<Integer>> combinationSum2(int[] num, int target) {

        if(num == null || num.length == 0)    return rst;

        List<Integer> list = new ArrayList<Integer>();

        Arrays.sort(num);

        helper(list, num, target, 0, 0);

        return rst;

    }

    private List<List<Integer>> rst = new ArrayList<List<Integer>>();

    public void helper(List<Integer> list, int[] candidates, int target, int sum, int start)

    {

        if(sum > target)    return;

        if(sum == target)

        {

            if(!rst.contains(list))

                rst.add(new ArrayList(list));

            return;

        }

        for(int i = start ; i < candidates.length; i++)

        {

            list.add(candidates[i]);

            helper(list, candidates, target, sum + candidates[i], i + 1);

            list.remove(list.size() - 1);

        }

    }

}

View Code

 学习之处：

　　集合中的元素都访问到了，所以时间复杂度为O(2ⁿ)，其实本质上就是一个二叉树