[array] leetcode - 40. Combination Sum II - Medium
阿新 • • 發佈:2017-11-16
可重復 one 實現 algorithm nat oos duplicate .cn ber
leetcode - 40. Combination Sum II - Medium
descrition
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.
- 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]
]
解析
和 leetcode - 39. Combination Sum - Medium 類似,只是這裏的數組元素存在重復,並且元素不可重復取。
代碼只實現了其中一種遞歸形式,這樣的實現方法遞歸層數應該是最淺的。
code
#include <iostream> #include <vector> #include <algorithm> using namespace std; class Solution{ public: vector<vector<int> > combinationSum2(vector<int> &candidates, int target){ vector<vector<int> > ans; vector<int> vecCur; sort(candidates.begin(), candidates.end()); combinationSum2Backtracking(candidates, 0, vecCur, target, ans); return ans; } void combinationSum2Backtracking(vector<int>& candidates, int index, vector<int>& vecCur, int target, vector<vector<int> > &ans){ if(target < 0) return; if(target == 0){ if(!vecCur.empty()) ans.push_back(vecCur); return; } for(int i=index; i<candidates.size(); i++){ if(candidates[i] > target) // candidates mush in ascending order break; // choose candidates[i], and each number in candidates may only // be used onece in combination vecCur.push_back(candidates[i]); combinationSum2Backtracking(candidates, i+1, vecCur, target - candidates[i], ans); // don‘t choose candidates[i] vecCur.pop_back(); // skip the duplicate while((i+1)<candidates.size() && candidates[i+1] == candidates[i]) i++; // after i++, i will point to a new unique number } } }; int main() { return 0; }
[array] leetcode - 40. Combination Sum II - Medium