LeetCode-Combination Sum

最新推荐文章于 2021-03-18 11:37:34 发布

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本文提供了一种解决LeetCode上组合总和问题的方法，采用深度优先搜索(DFS)算法，通过递归的方式找到所有可能的组合，确保结果无重复且元素按非递减顺序排列。

题目：https://oj.leetcode.com/problems/combination-sum/

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

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

Note:

All numbers (including target) will be positive integers.
Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

源码：Java版本
算法分析：时间复杂度O(n!)，空间复杂度O(n)

public class Solution {
   public List<List<Integer>> combinationSum(int[] candidates,
			int target) {
		if (candidates == null) {
			return null;
		}
		Set<List<Integer>> results = new HashSet<List<Integer>>();
		Stack<Integer> stack = new Stack<Integer>();
		Arrays.sort(candidates);
		DFS(candidates,0, target, stack, results);
		return new ArrayList<List<Integer>>(results);
	}

	@SuppressWarnings("unchecked")
	private void DFS(int[] candidates,int start, int target, Stack<Integer> stack,
			Set<List<Integer>> results) {
		if (target == 0) {
			results.add((Stack<Integer>) (stack.clone()));
		}
		for (int i = start; i < candidates.length; i++) {
			stack.push(candidates[i]);
			if (target - candidates[i] >= 0) {
				DFS(candidates, i,target - candidates[i], stack, results);
			}
			stack.pop();
		}
	}
    
}

代码解释：

1. 因为要升序，所有先对数组进行排序;

2. 因为要升序，所以下次遍历的开始位置是当前位置，所以添加了start变量；

3. 结果要无重复，故先用set收集结果去重。