Combination Sum II

最新推荐文章于 2025-05-25 18:32:47 发布

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本文介绍了一种寻找候选数字集合中所有唯一组合的方法，这些组合的数字之和等于目标数。该算法确保每个数字在组合中仅被使用一次，并通过深度优先搜索实现，同时采用排序和跳过重复值的技术来避免重复的解决方案。

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.

public class Solution {
    /**
     * @param num: Given the candidate numbers
     * @param target: Given the target number
     * @return: All the combinations that sum to target
     */
    public List<List<Integer>> combinationSum2(int[] num, int target) {
        List<List<Integer>> results = new ArrayList<>();
        if (num == null || num.length == 0) {
            return results;
        }
        Arrays.sort(num);
        List<Integer> result = new ArrayList<>();
        dfsHelper(num, target, 0, results, result);
        return results;
    }
    private void dfsHelper(int[] num,
                           int remainTarget,
                           int index,
                           List<List<Integer>> results,
                           List<Integer> result) {
        if (remainTarget == 0) {
            results.add(new ArrayList<>(result));
        }
        //不使用hash进行去重，if语句的作用是保证不跳跃着取相同的值，如 1,2',2''，不出现1,2''
        for (int i = index; i < num.length; i++) {
            if (i != index && num[i] == num[i - 1]) {
                continue;
            }
            if (remainTarget < num[i]) {
                break;
            }
            result.add(num[i]);
            dfsHelper(num, remainTarget - num[i], i + 1, results, result);
            result.remove(result.size() - 1);
        }
    }
}