Combination Sum II
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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 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
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C和一个特定值T,要求找出这里面满足以下条件的所有答案:数组中数字的值加起来等于特定和的答案.并且数组中的数字不允许被使用多次,但如果一开始就存在多个的话,可以使用多次。
分析:
可以采用与Combination Sum 类似的方法,但是由于题二中的每个数只能取一次,所以dfs函数中的for循环,i应从l+1开始,表示取下一个数。但这样带来的问题是,结果中会出现重复的取数方案,拿上面的例子来分析:C中有两个1可以选,那第一个1和7是一种可选方案(1+7=8),第二个1和7也是一种可选方案,按照上述算法,[1,7]会在结果中出现两次。当然可以对最后结果去重(如果用C++的话,sort->unique->erase可以实现)。不幸的是,这种解法会超时。解决超时的方案是不要将重复的方案加入到结果集中,也就避免了去重的工作。
<pre name="code" class="java">public class Solution { public List<List<Integer>> combinationSum2(int[] num, int target) { LinkedList<List<Integer>> res=new LinkedList<List<Integer>>(); LinkedList<Integer> list=new LinkedList<Integer>(); if(num==null || num.length<=0) return res; Arrays.sort(num); dfs(0,0,target,num,list,res); HashSet set=new HashSet(res); res.clear(); res.addAll(set); return res; } public void dfs(int start,int sum,int target,int[] num,LinkedList<Integer> list,List<List<Integer>> res){ if(sum==target){ LinkedList<Integer> l=new LinkedList<Integer>(list); res.add(l); return; } for(int i=start;i<num.length;i++){ if(sum+num[i]>target) return; list.add(num[i]); dfs(i+1,sum+num[i],target,num,list,res); list.pollLast(); } }}
注意:用set剔除重复的组。
c++:
class Solution {public: void internalCombinationSum2(vector<int> &num, int start, int sum, int target, vector<int> &combination, vector<vector<int> > &result) { int size = num.size(); if (sum == target) { result.push_back(combination); return; } else if ((start >= size) || (sum > target)) { return; } for (int i = start; i < size; ) { combination.push_back(num[i]); internalCombinationSum2(num, i + 1, sum + num[i], target, combination, result); combination.pop_back(); int j = i + 1; while (j < size) { if (num[i] == num[j]) { ++j; } else { break; } } i = j; } } vector<vector<int> > combinationSum2(vector<int> &num, int target) { vector<vector<int> > result; vector<int> combination; sort(num.begin(), num.end()); internalCombinationSum2(num, 0, 0, target, combination, result); return result; }};
http://www.cnblogs.com/panda_lin/archive/2013/11/05/combination_sum_ii.html
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