[Leetcode 40]组合数和II Combination Sum II

【题目】

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

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

Note:

- All numbers (including target) will be positive integers.
- The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,

A solution set is:

[

  [1, 7],

  [1, 2, 5],

  [2, 6],

  [1, 1, 6]

]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,

A solution set is:

[

  [1,2,2],

  [5]

]

【思路】

回溯，关键在去重：sort后nums[i-1]==nums[i] continue。

1、[Leetcode 78]求子集 Subset https://www.cnblogs.com/inku/p/9976049.html

2、[Leetcode 90]求含有重复数的子集 Subset II https://www.cnblogs.com/inku/p/9976099.html

3、讲解在这： [Leetcode 216]求给定和的数集合 Combination Sum III

4、[Leetcode 39]组合数的和Combination Sum

【代码】

有参考39的思路设置全局变量+两种情况合并。

class Solution {

    private static List<List<Integer>> ans ;

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

        Arrays.sort(candidates);

        ans = new ArrayList<>();

        fun(new ArrayList<Integer>(),candidates,target,0);

        return ans;

    }

    public void fun(List<Integer> tmp,int[] data, int aim,int flag){

        if(aim<=0){

            if(aim==0){

                ans.add(new ArrayList<>(tmp));

            }

            return;

        }

        for(int i=flag;i<data.length;i++){

            if(i>flag&&data[i-1]==data[i])

                continue;

            tmp.add(data[i]);

            fun(tmp,data,aim-data[i],i+1);

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

        }

    }

}

巴特西

[Leetcode 40]组合数和II Combination Sum II

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