【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, … , 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] 

解答

题目给出一个候选数列和一个目标数值，找出所有相加等于目标数值的数列，原候选数列的元素可以重复。

NP问题，先排好序，然后每次递归中把剩下的元素一一加到结果集合中，并且把目标减去加入的元素，然后把剩下元素（包括当前加入的元素）放到下一层递归中解决子问题。代码如下：

public class Solution {    public List<List<Integer>> combinationSum(int[] candidates, int target) {        List<List<Integer>> res=new ArrayList<List<Integer>>(); //注意多维List的初始化        if(candidates==null||candidates.length==0){        return res;        }        Arrays.sort(candidates);        helper(candidates,0,target,new ArrayList<Integer>(),res);        return res;    }    private void helper(int[] candidates,int start,int target,ArrayList<Integer> item,List<List<Integer>> res){ //注意和前面类型对应    if(target<0){    return;    }    if(target==0){    res.add(new ArrayList<Integer>(item));    return;    }    for(int i=start;i<candidates.length;i++){    if(i>0&&candidates[i]==candidates[i-1]){    continue;    }    item.add(candidates[i]);    helper(candidates,i,target-candidates[i],item,res);    item.remove(item.size()-1);    }    }}

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

---EOF---