leetcode--Combination Sum

1.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]

2.解法分析

实际上这是一个类似于深度搜索的算法，代码如下，

1 class Solution {

2 public:

3     vector<vector<int> > combinationSum(vector<int> &candidates, int target) {

4         // Start typing your C/C++ solution below

5         // DO NOT write int main() function

6         vector<int>cur;

7         vector<vector<int>>result;

8         int sum=0;

9         sort(candidates.begin(),candidates.end());

10         myCombinationSum(candidates,cur,result,sum,0,target);

11

12         return result;

13

14     }

15

16     void myCombinationSum(vector<int> &candidates,vector<int>&cur,vector<vector<int>>&result,int &curSum,int start,int target)

17     {

18         if(start>=candidates.size())return ;

19         for(int i=start;i<candidates.size();++i)

20         {

21             curSum+=candidates[i];

22             cur.push_back(candidates[i]);

23             if(curSum==target)

24             {//搜索到了满足要求的一条路径，那么按照之前的路径继续下移或右移肯定得不到另外满足的路径，因此只能上移

25                 result.push_back(cur);curSum-=candidates[i];

26                 cur.pop_back();break;

27             }

28             else

29                 if(curSum<target)

30                 {

31                     myCombinationSum(candidates,cur,result,curSum,i,target);

32                     curSum-=candidates[i];

33                     cur.pop_back();

34                 }

35                 else

36                 {

37                     curSum-=candidates[i];

38                     cur.pop_back();

39                     break;

40                 }

41         }

42

43     }

44 };