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[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
10,1,2,7,6,1,5and target8,
A solution set is:[1, 7][1, 2, 5][2, 6][1, 1, 6]
以下是我 AC 的代码：
/** * author: Zhou J */class Solution {public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { sort(num.begin(), num.end()); vector<vector<int>> ret; vector<int> path; combinationSum2Rec(num, target, ret, path, 0); return ret; } private: void combinationSum2Rec(const vector<int> &num, const int target, vector<vector<int>> &ret, vector<int> &path, size_t pos) { if (target == 0) { ret.push_back(path); return; } // Mark the previous one int prev = -1; for (size_t ix = pos; ix != num.size(); ++ix) { if (num[ix] > target) { break; } if (prev == num[ix]) { continue; } path.push_back(num[ix]); combinationSum2Rec(num, target - num[ix], ret, path, ix + 1); // Backtracing. Note that we should sort the vector before we use the skill of backtracing. path.erase(path.end() - 1); // The one which we have pushed just now. prev = num[ix]; } }};
[LeetCode] Combination Sum II
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