LeetCode_OJ【40】Combination Sum II

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]

本题和39题类似，只不过数组中的数字不能重复选取，而且数组中可能会有重复的数字，导致结果集中会出现重复解，因此如何过滤掉重复解也是本题的关键。

下面是本题的JAVA实现。

public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> res = new ArrayList<List<Integer>>();
List<Integer> tmp = new ArrayList<Integer>();
dfs(res,tmp,candidates,target,0);
return res;
}

public void dfs(List<List<Integer>> res,List<Integer> tmp,int[] candidates, int gap,int index){
if(gap == 0){
List<Integer> list = new ArrayList<Integer>(tmp);
res.add(list);
return;
}
for(int i = index;i < candidates.length; i++){
if(gap < candidates[i])
return;
else{
tmp.add(candidates[i]);
dfs(res,tmp,candidates,gap - candidates[i],i +1);
tmp.remove(tmp.size() -1);
while(i + 1 < candidates.length && candidates[i] == candidates[i +1])
i++;
}
}
}
}