题目描述:
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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
给定一个数组,和一个目标数,使用给定数组中的数字,求出所有和等于目标数的可能组合。 例如给定2,3,6,7 ,可能组合是[7]和[2,2,3]
这个题目有两个条件:
1. 数字是可以重复的
2. 组合必须是升序排列
3. 不能有重复组合
4. 所有数字都是正数
思路:
对数组遍历,缩小目标数: target-= candidate[i]。如果target < candadite[i] ,中断循环
使用1个数组:arr记录当前遍历情况,如果target为0,存入结果。
在遍历数组过程中,添加当前元素:arr.Add(self),进入递归
拿掉当前元素: arr.Remove(self)
实现代码:
public IList<IList<int>> CombinationSum(int[] candidates, int target)
{
if(candidates == null || candidates.Length == 0){
return null;
}
var arr = candidates.OrderBy(x=>x).ToList();
IList<IList<int>> result = new List<IList<int>>();
Travel(arr ,new List<int>(), 0, target, result);
return result;
}
private void Travel(IList<int> candidates, IList<int> arr, int index, int target, IList<IList<int>> result){
if(target == 0 ){
result.Add(new List<int>(arr));
return ;
}
for(var i = index ;i < candidates.Count; i++){
if(target < candidates[i]){
return;
}
arr.Add(candidates[i]);
Travel(candidates, arr, i + 1 , target - candidates[i], result);
arr.Remove(candidates[i]);
}
}
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LeetCode – Combination Sum
原文:http://blog.csdn.net/lan_liang/article/details/49108349