[leetcode]40. Combination Sum II组合之和(每元素限用一次)

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]
]

 

题意：

给定一个集合以及一个值target，找出所有加起来等于target的组合。（每个元素只能用一次）

 

Solution1： Backtracking

在[leetcode]39. Combination Sum组合之和的基础上， 加一行查重的动作。

 

code

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        helper(candidates, 0, target, path, result );
        return result;
    }
    
    private void helper(int[] nums, int index, int remain, List<Integer> path, List<List<Integer>> result){
        if (remain == 0){
            result.add(new ArrayList<>(path));
            return;
        }  
        for(int i = index; i < nums.length; i++){
            if (remain < nums[i]) return;
            if(i > index && nums[i] == nums[i-1]) continue; /** skip duplicates */
            path.add(nums[i]);
            helper(nums, i + 1, remain - nums[i], path, result);
            path.remove(path.size() - 1);
        }
    }
}