[leetcode] 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 set2,3,6,7and target7,
A solution set is:
[7]

[2, 2, 3]

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

思路：返回所有可能，只能枚举了，先排序，然后此题每个元素可以取多次，所以递归下一层的时候选取的元素不变（区别Combination Sum II 中递归下一层只能从后面元素选取）。

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

public class Solution {
	public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates,
			int target) {
		ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
		if (candidates == null || candidates.length == 0)
			return result;

		int n = candidates.length;

		Arrays.sort(candidates);
		ArrayList<Integer> list = new ArrayList<Integer>();

		dfs(0, candidates, target, list, result);

		return result;
	}

	private void dfs(int level, int[] a, int num, ArrayList<Integer> list,
			ArrayList<ArrayList<Integer>> result) {
		if (num == 0) {
			result.add(new ArrayList<Integer>(list));
		} else if (num < 0)
			return;
		else {
			for (int i = level; i < a.length; i++) {
				if (a[i] <= num) {
					list.add(a[i]);
					dfs(i, a, num - a[i], list, result);
					list.remove(list.size() - 1);
				}
			}

		}
	}

	public static void main(String[] args) {
		System.out.println(new Solution().combinationSum(
				new int[] { 2, 3, 6, 7 }, 7));
		System.out.println(new Solution().combinationSum(
				new int[] { 1, 3, 6, 7 }, 7));

		System.out.println(new Solution().combinationSum(
				new int[] { 7,3,2 }, 18));
	}

}

第二遍记录：

注意去重的方法，先排序，然后判断，类似permutation的去重，if (i == start || candidates[i] != candidates[i - 1])

注意新的start的设置。  

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>> res = new ArrayList<List<Integer>>();
        if (candidates == null || candidates.length == 0)
            return res;
        Arrays.sort(candidates);
        List<Integer> tmp = new ArrayList<Integer>();
        combine(candidates, res, tmp, target, 0);
        return res;
    }

    private void combine(int[] candidates, List<List<Integer>> res, List<Integer> tmp, int target, int start) {
        if (target < 0)
            return;
        if (target == 0) {
            res.add(new ArrayList<Integer>(tmp));
            return;
        } else {
            for (int i = start; i < candidates.length; i++) {
                //注意这步的去重
                if (i == start || candidates[i] != candidates[i - 1]) {
                    tmp.add(candidates[i]);
                    //注意新的start是i,表示只能从以选中元素及其后面玄素开始选。
                    combine(candidates, res, tmp, target - candidates[i], i);
                    tmp.remove(tmp.size() - 1);
                }
            }

        }
    }

    public static void main(String[] args) {
        System.out.println(new Solution().combinationSum(new int[] { 2, 3, 6, 7 }, 7));
        System.out.println(new Solution().combinationSum(new int[] { 1, 1, 1, 2, 2 }, 10));
    }

}

第三遍： 注意去重。 

参考：

http://blog.csdn.net/linhuanmars/article/details/20828631