Combination Sum | && || Leetcode

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] 

這個題目其實就是經典回溯問題，但是就是有一個不太一樣地方就是，這裏每個數可以出現無限制次數，加入說每個數只出現一次就跟這道題一樣啦：

http://blog.csdn.net/huruzun/article/details/21823343

遞歸調用時需要增加一個控制起始位置參數來使得每次還可以調用出現過的數

package combination_Sum;

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


public class Solution {
	public static List<List<Integer>> combinationSum(int[] candidates, int target) {
	    List<List<Integer>> result = new ArrayList<>();
	 
	    if(candidates == null || candidates.length == 0) 
	    	return result;
	 
	    ArrayList<Integer> current = new ArrayList<Integer>();
	    Arrays.sort(candidates);
	    
	    combinationSum(candidates, target, 0, current, result);
	 
	    return result;
	}
	 
	public static void combinationSum(int[] candidates, int target, int j, ArrayList<Integer> curr, List<List<Integer>> result){
	   if(target == 0){
		   // 這裏一定要new起來一個新的拷貝
	       ArrayList<Integer> temp = new ArrayList<Integer>(curr);
	       result.add(temp);
	       return;
	   }
	   // 這裏 參數 j 用來控制同一個數可以出現多次
	   for(int i=j; i<candidates.length; i++){
	       if(target < candidates[i]) 
	            return;
	       
	       curr.add(candidates[i]);
	       combinationSum(candidates, target - candidates[i], i, curr, result);
	       curr.remove(curr.size()-1); 
	   }
	}
	public static void main(String[] args){
		int []candidates = {2,3,6,7};
		System.out.println(combinationSum(candidates,7));
	}
}

接着是這道題的變形

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

與前面區別就是每個數字只能使用一次，而且要避免重複的答案，最簡單就是在最終答案處進行判重，但是這樣效率不高，在對數組排序後，如果某個數字後面又重複的就需要跳過。

public class Solution {
   	public static List<List<Integer>> combinationSum2(int[] candidates,
			int target) {
		List<List<Integer>> result = new ArrayList<>();

		if (candidates == null || candidates.length == 0)
			return result;

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

		combinationSum(candidates, target, 0, current, result);

		return result;
	}

	public static void combinationSum(int[] candidates, int target, int i,
			ArrayList<Integer> curr, List<List<Integer>> result) {
		if (target == 0) {
			// 這裏一定要new起來一個新的拷貝
			
			ArrayList<Integer> temp = new ArrayList<Integer>(curr);
			
			result.add(temp);
			return;
		}
		for (; i < candidates.length; i++) {
			if (target < candidates[i])
				return;
			
			curr.add(candidates[i]);
			combinationSum(candidates, target - candidates[i], i+1, curr, result);
			curr.remove(curr.size() - 1);
			// 這裏重複的數字要跳過
			while(i< candidates.length-1 && candidates[i] == candidates[i+1]){
				i++;
			}
		}
	}
	
}