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
,
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
类似与之前的Combination Sum的DFS,有一点需要注意,如何避免重复。如果两个数相同,我们先用前一个数,只有当前一个数用了,这个数才能使用。
例如:1 1。
当我们要使用第二个1时,我们要检查他的前面一个1是否使用了,当未被使用时第二个1就不能使用。
1 class Solution { 2 private: 3 vector> ret; 4 vector a; 5 public: 6 void solve(int dep, int maxDep, vector &num, int target) 7 { 8 if (target < 0) 9 return;10 11 if (dep == maxDep)12 {13 if (target == 0)14 {15 vector res;16 for(int i = 0; i < maxDep; i++)17 for(int j = 0; j < a[i]; j++)18 res.push_back(num[i]);19 ret.push_back(res);20 }21 22 return;23 }24 25 for(int i = 0; i <= min(target / num[dep], 1); i++)26 {27 a[dep] = i;28 29 if (i == 1 && dep > 0 && num[dep-1] == num[dep] && a[dep-1] == 0)30 continue;31 32 solve(dep + 1, maxDep, num, target - i * num[dep]);33 }34 }35 36 vector > combinationSum2(vector &num, int target) {37 // Start typing your C/C++ solution below38 // DO NOT write int main() function39 sort(num.begin(), num.end());40 a.resize(num.size());41 ret.clear();42 if (num.size() == 0)43 return ret;44 45 solve(0, num.size(), num, target);46 47 return ret;48 }49 };