LeetCode-Combination Sum II-优快云博客

本文讨论了如何在给定候选数集合和目标数的情况下，找出所有独特的组合方式，使得这些组合的元素之和等于目标数。文章强调了算法的时间复杂度为O(n!)，空间复杂度为O(n)，并提供了Java实现代码。

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

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.
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 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

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>> combinationSum2(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+1,target - candidates[i], stack, results);
			}
			stack.pop();
		}
	}
}