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Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations inC 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]
答案
public class Solution {
int[] candidates;
public List<List<Integer>> combinationSum(int index ,int target)
{
List<List<Integer>> result=new LinkedList<List<Integer>>();
if(index>=candidates.length||candidates[index]>target)
{
return result;
}
List<Integer> p=new LinkedList<Integer>();
int end;
for(end=index;end<candidates.length;end++)
{
if(candidates[end]!=candidates[index])
{
break;
}
}
int times=end-index;
for(int time=0;time<=times;time++)
{
if(target==0)
{
result.add(p);
break;
}
List<List<Integer>> next=combinationSum(end,target);
if(next.size()>0)
{
for(List<Integer> list:next)
{
List<Integer> pList=new LinkedList<Integer>();
pList.addAll(p);
pList.addAll(list);
result.add(pList);
}
}
p.add(candidates[index]);
target-=candidates[index];
}
return result;
}
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> result=new LinkedList<List<Integer>>();
if(candidates==null||target<=0)
{
return result;
}
Arrays.sort(candidates);
this.candidates=candidates;
return combinationSum(0,target);
}
}Combination Sum II
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