Leetcode | Combination Sum I && II

 Combination Sum 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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

先排序，然后每次从当前位置开始找下一个数。去下重，不过貌似leetcode的testcases没有针对这个。

 1 class Solution {

 2 public:

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

 4         vector<vector<int> > ans;

 5         if (candidates.empty()) return ans;

 6         vector<int> tmp;

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

 8         recurse(candidates, 0, target, tmp, ans);

 9         return ans;

10     }

11     void recurse(vector<int> &candidates, int start, int target, vector<int> &tmp, vector<vector<int> > &ans) {

12         if (target < 0) return;

13         if (target == 0) {

14             ans.push_back(tmp);

15             return;

16         }

17         

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

19             tmp.push_back(candidates[i]);

20             recurse(candidates, i, target - candidates[i], tmp, ans);

21             tmp.pop_back();

22         }

23     }

24 };

 

 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]

这道题和上面类似。但是每个数只能使用一次。如果是重复数，在每个位置上用的应该是第一个数，这样才能保证这个重复数可以重复使用。

比如[1,1,2,3]，如果每个位置用的是重复数的最后一个，那么就生成不了[1,1,2]了。

 1 class Solution {

 2 public:

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

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

 5         vector<int> r;

 6         recursive(num, target, 0, r);

 7         return ret;

 8     }

 9     

10     void recursive(vector<int> &num, int target, int start, vector<int> &r) {

11         if (target <= 0) {

12             if (0 == target) ret.push_back(r);

13             return;

14         }

15         

16         for (int i = start; i < num.size(); ++i) {

17             if (i > start && num[i] == num[i - 1]) continue;

18             r.push_back(num[i]);

19             recursive(num, target - num[i], i + 1, r);

20             r.pop_back();

21         }

22     }

23 private:

24     vector<vector<int> > ret;

25 };