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 };