【Leetcode】Combination sum 1，2

【題目】

【分析】

首先，先sort,

如果target > 0:

【代碼】

using recursive 這恐怕的 O(n2)...

過程：

before recursion : [2]

before recursion : [2, 2]

before recursion : [2, 2, 2]

after recursion and remove : [2, 2]

before recursion : [2, 2, 3]

after recursion and remove : [2, 2]

after recursion and remove : [2]

before recursion : [2, 3]

after recursion and remove : [2]

after recursion and remove : []

before recursion : [3]

before recursion : [3, 3]

after recursion and remove : [3]

after recursion and remove : []

before recursion : [6]

after recursion and remove : []

before recursion : [7]

after recursion and remove : []

[[2, 2, 3], [7]]

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

        return result;
    }

    private void getResult(List<List<Integer>> result, List<Integer> cur, int candidates[], int target, int start){
        if(target > 0){
            for(int i = start; i < candidates.length && target >= candidates[i]; i++){
                cur.add(candidates[i]);
                getResult(result, cur, candidates, target - candidates[i], i);
                cur.remove(cur.size() - 1);
            }//for
        }//if
        else if(target == 0 ){
            result.add(new ArrayList<Integer>(cur));
        }//else if
    }
}

【二】

【題目】

Given a collection of candidate numbers (C) and a target number (T)，find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
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 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

【題目】

題意分析：從給定數組中找到一組數字，要求這組數字之和等於target。另外，數組中的數字不允許被使用多次，但如果一開始就存在多個的話，可以使用多次。
解題思路：顯然先排序，然後dfs。其中有一點要注意的是：因爲不能重複，所以要跳過一樣的數字。以上面爲例，如果不跳過重複的1的話，會出現多個：[1,7]

	    public static void dfs(int [] num, int start, int target, List<Integer> array, List<List<Integer>> result) {  
	        if(target==0) {  
	            result.add(new ArrayList<Integer>(array));  
	            return;  
	        }  
	          
	        if(start>=num.length||num[0]>target) {  
	            return;  
	        }  
	        int i = start;  
	        while(i<num.length) {  
	            if(num[i]<=target) {  
	                array.add(num[i]);  System.out.println("before dfs: " + array);
	                dfs(num, i + 1, target - num[i], array, result);  
	                array.remove(array.size()-1);  System.out.println("after dfs and rempve: " + array);
	                //跳過重複的元素  
	                while(i<(num.length-1)&&num[i]==num[i+1]) {  
	                    i++;  
	                }  
	            }  
	            i++;  
	        }  
	    }  
	      
	      
	    public static List<List<Integer>> combinationSum2(int[] num, int target) {  
	         List<List<Integer>> result = new ArrayList<List<Integer>>();  
	         ArrayList<Integer> array = new ArrayList<Integer>();  
	         if(num==null) {  
	             result.add(array);  
	             return result;  
	         }  
	         Arrays.sort(num);  
	         dfs(num,0, target,array,result);   
	         return result;  
	    }