LeetCode | Combination Sum

题目

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] 

用递归的动态规划解决，可以加上备忘录，如果target值允许的话，可以凶残的用非递归的数组方式解决。

代码

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

public class CombinationSum {
	public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates,
			int target) {
		Arrays.sort(candidates);
		ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
		ArrayList<Integer> list = new ArrayList<Integer>();
		solve(results, list, candidates, 0, target);
		return results;
	}

	private void solve(ArrayList<ArrayList<Integer>> results,
			ArrayList<Integer> list, int[] candidates, int i, int target) {
		if (target < 0) {
			return;
		}
		if (target == 0) {
			results.add(new ArrayList<Integer>(list));
			return;
		}
		for (int j = i; j < candidates.length && candidates[j] <= target; ++j) {
			list.add(candidates[j]);
			solve(results, list, candidates, j, target - candidates[j]);
			list.remove(list.size() - 1);
		}
	}
}

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LeetCode | Combination Sum

原文：http://blog.csdn.net/perfect8886/article/details/22689327