description:

給定target， 求給定數列中找到幾個數（其中的數不可以重復使用，且一組數有幾個也不做限制）的和為target，和上面那個題一毛一樣的，就改一下下標就行了，背下來背下來背下來~~~
Note:

All numbers (including target) will be positive integers. The solution set must not contain duplicate combinations. //這個條件注意了，只要是兩個挨著的一樣，那么就會導致最后出現相同的組合

Example:

Example 1:Input: candidates = [10,1,2,7,6,1,5], target = 8, A solution set is: [[1, 7],[1, 2, 5],[2, 6],[1, 1, 6] ]Example 2:Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [[1,2,2],[5] ]

answer:

class Solution { public:vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {vector<vector<int>> res;vector<int> out;sort(candidates.begin(), candidates.end()); //這里剛開始就沒寫，一定要先變成有序數組！！！combinationDFS(candidates, res, out, target, 0);return res;}void combinationDFS(vector<int>& candidates, vector<vector<int>>& res, vector<int>& out, int target, int start) {if(target < 0) return;if(target == 0) {res.push_back(out);return;}for(int i = start; i < candidates.size(); i++) {if(i > start && candidates[i] == candidates[i - 1]) continue; //這里也要注意了，因為上邊那個note的第二條out.push_back(candidates[i]);combinationDFS(candidates, res, out, target - candidates[i], i + 1);out.pop_back();}} };

relative point get√：

hint :

轉載于:https://www.cnblogs.com/forPrometheus-jun/p/11166218.html

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