LeetCode Combination Sum II

题目：

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where
the candidate numbers sums toT.

Each number in C may only be used once in the combination.

Note:

All numbers (including target) will be positive integers.
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]
]

题意：
给定一个数组和一个target，那么要求在这一数组中寻找任意的组合，使其和等于给定的target，遇到好几组相同的，要去重。那么也是考虑用DFS来做，结合递归的思路。此题与上一题思路一致，我们都可以采用相同的思路，只是每一次内层递归结束，退出到外层时，等在原始数组中对下一个数组元素进行判断，看它是不是与之前的那个元素一致，如果相同，那么指针继续往后面走。这里是需要注意的。

public class combinationSum2
{
public List<List<Integer>> combinationSum2(int[] candidates,int target)
{
List<List<Integer>> list = new ArrayList<List<Integer>>();
ArrayList<Integer> l = new ArrayList<Integer>();
int length = candidates.length;
if(length == 0 || list == null)
return list;
Arrays.sort(candidates);
dfs(candidates,target,0,length,0,list,l);
return list;
}
public void dfs(int[] nums,int target,int start,int n,int sum,List<List<Integer>> list,ArrayList<Integer> l)
{
if(sum == target)
{
list.add(new ArrayList<Integer> (l));
return;
}
if(sum > target)
return;
for(int j = start; j < n; j++)
{
l.add(nums[j]);
dfs(nums,target,j + 1,n,sum + nums[j],list,l);
l.remove(l.size() - 1);
while(j < n -1)
{
if(nums[j] == nums[j + 1])
j++;
else
break;
}
}
}
}