Leetcode 39 Combination Sum

最新推荐文章于 2024-11-20 10:20:22 发布

原创最新推荐文章于 2024-11-20 10:20:22 发布 · 267 阅读

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本文深入探讨了如何使用回溯DFS方法解决组合总和问题，即在无重复候选数集中找到所有可能的组合，使这些组合的数总和等于目标数。文章提供了详细的算法实现过程及示例代码。

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Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

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

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

这个题就是典型的回溯dfs方法

1）

class Solution {
    public List<List<Integer>> combinationSum(int[] nums, int target) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums.length == 0) {
            return result;
        }
        List<Integer> path = new ArrayList<>();
        dfs(nums, target, result, path, 0);
        return result;
    }
    private void dfs(int[] nums, int target, List<List<Integer>> result, List<Integer> path, int index) {
        if (target < 0) {
            return;
        }
        if (target == 0) {
            result.add(new ArrayList<>(path));
        }
        for (int i = index; i < nums.length; i++) {
            path.add(nums[i]);
            dfs(nums, target - nums[i], result, path, i);
            path.remove(path.size() - 1);
        }
    }
    
}