• 39. Combination Sum


    Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

    The same repeated number may be chosen from candidates unlimited number of times.

    Note:

    • All numbers (including target) will be positive integers.
    • The solution set must not contain duplicate combinations.

    Example 1:

    Input: candidates = [2,3,6,7], target = 7,
    A solution set is:
    [
      [7],
      [2,2,3]
    ]
    

    Example 2:

    Input: candidates = [2,3,5], target = 8,
    A solution set is:
    [
      [2,2,2,2],
      [2,3,3],
      [3,5]
    ]


    考察要点: 基本的DFS, 其他没有难度, 注意数组的元素可以被重复利用
    class Solution {
    public:
        vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
            sort(candidates.begin(),candidates.end());
            vector<vector<int>> res;
            vector<int> candi;
            traverse(candidates,target,res,candi,0);
            return res;
        }
        
        void traverse(vector<int> &candidates,int target, vector<vector<int>> &res, vector<int> &cur, int index)
        {
            if(0==target)
            {
                res.push_back(cur);
                return;
            }
            
            for(;index<candidates.size()&&target>=candidates[index];++index)
            {
                cur.push_back(candidates[index]);
                traverse(candidates,target-candidates[index],res,cur,index);
                cur.pop_back();
            }
        }
    };
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  • 原文地址:https://www.cnblogs.com/lychnis/p/11704868.html
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