Leetcode 40. Combination Sum II (Medium) (java)

最新推荐文章于 2025-07-29 14:24:32 发布

Niko_Ke

最新推荐文章于 2025-07-29 14:24:32 发布

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本文介绍了解决LeetCode第40题“组合总和 II”的方法，该题要求从候选数字集合中找出所有不重复的组合，使组合内的数字之和等于目标值。文章详细解释了如何使用回溯算法实现这一功能，并通过示例展示了正确答案。

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Leetcode 40. Combination Sum II (Medium) (java)

Tag: Array, Backtracking

Difficulty: Medium

/*

40. Combination Sum II (Medium)

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.
The solution set must not contain duplicate combinations.
For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8, 
A solution set is: 
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

*/
public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(candidates);
        combinationSum2(candidates, target, 0, res, new ArrayList<>());
        return res;
    }
    private void combinationSum2(int[] candidates, int target, int start, List<List<Integer>> res, List<Integer> res_sub) {
        if (start > candidates.length || target < 0) {
            return;
        } else if (target == 0) {
            res.add(new ArrayList(res_sub));
            return;
        }
        for (int i = start; i < candidates.length; i++) {
            if (i > start && candidates[i] == candidates[i - 1]) {
                continue;
            }
            res_sub.add(candidates[i]);
            combinationSum2(candidates, target - candidates[i], i + 1, res, res_sub);
            res_sub.remove(res_sub.size() - 1);
        }
    }
}