leetcode_Combination Sum
2015-08-07 10:32
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描述:
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
A solution set is:
思路:
这种题目,一般要用到递归或回溯两种方法,用回溯法试过,代码规模总是越来越庞大,但最终还是没能通过所有的测试用例,^-^!
用递归的话这题目看着要容易理解的多,每递归一次target要变为target=target-candidates[i],并将开始index赋值为i,当target==0时,条件满足,如果target<candidates[i],这轮循环结束,方法出栈,当前的i++,继续循环。
总之,用递归来解决问题,难想到,也不太容易被看懂。
代码:
public List<List<Integer>> combinationSum(int[] candidates, int target)
{
List<List<Integer>>listResult=new ArrayList<List<Integer>>();
ArrayList<Integer>list=new ArrayList<Integer>();
Arrays.sort(candidates);
combination(listResult, list, candidates, target,0);
return listResult;
}
public void combination(List<List<Integer>>listResult,
ArrayList<Integer>list,int[] candidates, int target,int begin)
{
if(target==0)
{
listResult.add(list);
return;
}
for(int i=begin;i<candidates.length&&target>=candidates[i];i++)
{
ArrayList<Integer>copy=new ArrayList<Integer>(list);
copy.add(candidates[i]);
combination(listResult, copy, candidates,target- candidates[i],i);
}
}
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
2,3,6,7and target
7,
A solution set is:
[7]
[2, 2, 3]
思路:
这种题目,一般要用到递归或回溯两种方法,用回溯法试过,代码规模总是越来越庞大,但最终还是没能通过所有的测试用例,^-^!
用递归的话这题目看着要容易理解的多,每递归一次target要变为target=target-candidates[i],并将开始index赋值为i,当target==0时,条件满足,如果target<candidates[i],这轮循环结束,方法出栈,当前的i++,继续循环。
总之,用递归来解决问题,难想到,也不太容易被看懂。
代码:
public List<List<Integer>> combinationSum(int[] candidates, int target)
{
List<List<Integer>>listResult=new ArrayList<List<Integer>>();
ArrayList<Integer>list=new ArrayList<Integer>();
Arrays.sort(candidates);
combination(listResult, list, candidates, target,0);
return listResult;
}
public void combination(List<List<Integer>>listResult,
ArrayList<Integer>list,int[] candidates, int target,int begin)
{
if(target==0)
{
listResult.add(list);
return;
}
for(int i=begin;i<candidates.length&&target>=candidates[i];i++)
{
ArrayList<Integer>copy=new ArrayList<Integer>(list);
copy.add(candidates[i]);
combination(listResult, copy, candidates,target- candidates[i],i);
}
}
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