【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

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