LeetCode 39. Combination Sum（組合求和）

原題網址：https://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, … , 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]

方法一：深度優先搜尋。

public class Solution {
    private List<List<Integer>> combinations = new ArrayList<>();
    
    private List<Integer> transfer(int[] candidates, List<Integer> nums) {
        List<Integer> result = new ArrayList<>();
        for(int i=0; i<nums.size(); i++) {
            for(int j=0; j<nums.get(i); j++) {
                result.add(candidates[i]);
            }
        }
        return result;
    }
    
    private void find(int[] candidates, List<Integer> nums, int sum) {
        int step = nums.size();
        if (step == candidates.length-1) {
            if (sum % candidates[step] == 0) {
                nums.add(sum / candidates[step]);
                combinations.add(transfer(candidates, nums));
                nums.remove(nums.size()-1);
            }
            return;
        }
        for(int i=0; i<=sum/candidates[step]; i++) {
            nums.add(i);
            find(candidates, nums, sum-candidates[step]*i);
            nums.remove((int)(nums.size()-1));
        }
    }
    
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        find(candidates, new ArrayList<Integer>(), target);
        return combinations;
    }
}
           

方法二：在方法一的基礎上優化，深度優先搜尋。

public class Solution {
    private void find(int[] candidates, int step, int sum, List<Integer> nums, List<List<Integer>> results) {
        if (sum == 0) {
            results.add(new ArrayList<>(nums));
            return;
        }
        if (sum < 0 || step >= candidates.length || candidates[step] > sum) return;
        nums.add(candidates[step]);
        find(candidates, step, sum-candidates[step], nums, results);
        nums.remove(nums.size()-1);
        find(candidates, step+1, sum, nums, results);
    }

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