LeetCode : Combination Sum [java]-CSDN博客

本文链接：https://blog.csdn.net/hackeraaa/article/details/50884760

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, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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]

思路：回溯法，为了保证重复使用数字，注意i每次从当前值取。

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Solution {
	public List<List<Integer>> combinationSum(int[] candidates, int target) {
		List<List<Integer>> result = new ArrayList<List<Integer>>();
		List<Integer> current = new ArrayList<Integer>();
		Arrays.sort(candidates);
		solve(candidates, target, 0, current, result);
		return result;
	}

	public void solve(int[] candidates, int target, int index, List<Integer> current, List<List<Integer>> result) {
		if (target == 0) {
			List<Integer> tmp = new ArrayList<Integer>(current);
			result.add(tmp);
			return;
		}
		for (int i = index; i < candidates.length; i++) {
			if (target < candidates[i]) {
				return;
			}
			current.add(candidates[i]);
			solve(candidates, target - candidates[i], i, current, result);
			current.remove(current.size() - 1);
		}
	}
}