[Leetcode] Combination Sum II

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.

 

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] 

记录当前idx之前的元素在结果中是否被使用，如果之前的与当前元素相等的元素没有被使用的话，那么这个元素也不应该被使用，这样就可以去重了。比如[1, 1, 2]，如果第一个1没有被使用，那么第二个1也不能使用。

 1 class Solution {

 2 public:

 3     void findNext(vector<int> &num, int target, vector<vector<int> > &res, vector<int> &v, int idx, int sum, vector<bool> &flag) {

 4         if (sum > target || idx > num.size()) return;

 5         if (sum == target) {

 6             res.push_back(v);

 7             return;

 8         }

 9         bool fflag = false;

10         for (int i = idx-1; i >= 0; --i) {

11             if (num[i] == num[idx] && !flag[i]) {

12                 fflag = true;

13                 break;

14             };

15             if (num[i] != num[idx]) break;

16         }

17         if (!fflag) {

18         v.push_back(num[idx]);

19             flag[idx] = true;

20             findNext(num, target, res, v, idx+1, sum+num[idx], flag);

21             flag[idx] = false;

22             v.pop_back();

23         }

24         findNext(num, target, res, v, idx+1, sum, flag);

25     }

26     

27     vector<vector<int> > combinationSum2(vector<int> &num, int target) {

28         vector<vector<int> > res;

29         vector<bool> flag(num.size(), false);

30         vector<int> v;

31         sort(num.begin(), num.end());

32         findNext(num, target, res, v, 0, 0, flag);

33         return res;

34     }

35 };

直接找的话会有重复的答案，可以先把结果存在一个set里，最然把结果从set转存到vector里。代码如下。

 1 class Solution {

 2 public:

 3     void findNext(vector<int> &num, int target, set<vector<int> > &res, vector<int> &v, int idx, int sum) {

 4         if (sum > target || idx > num.size()) return;

 5         if (sum == target) {

 6             res.insert(v);

 7             return;

 8         }

 9         v.push_back(num[idx]);

10         findNext(num, target, res, v, idx+1, sum+num[idx]);

11         v.pop_back();

12         findNext(num, target, res, v, idx+1, sum);

13     }

14     

15     vector<vector<int> > combinationSum2(vector<int> &num, int target) {

16         set<vector<int> > res;

17         vector<int> v;

18         sort(num.begin(), num.end());

19         findNext(num, target, res, v, 0, 0);

20         vector<vector<int> > ret;

21         for (set<vector<int> >::iterator i = res.begin(); i != res.end(); ++i) {

22             ret.push_back(*i);

23         }

24         return ret;

25     }

26 };