39. Combination Sum

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

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

Note:

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

For example, given candidate set [2, 3, 6, 7] and target 7, 
A solution set is: 

[
  [7],
  [2, 2, 3]
]

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来源： https://leetcode.com/problems/combination-sum/

分析

需要使用DFS来做，

[a, b, c, d, e, f, ...]

假设目标为f，数字从小到大排列，那么需要递归到数字f为止。

root(sum:f)           a(sum:f-a)

|                     |

a b c d e choose a => a b c d e 

  

终止条件：

1 找到结果

2 数组C中，最小的数字比剩余的数字大

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class Solution { public:     vector<vector<int>> combinationSum(vector<int>& candidates, int target) {         vector<vector<int>> results;         sort(candidates.begin(), candidates.end());         helper(results,vector<int>{},candidates,target,0);         return results;     }     void helper(vector<vector<int>> &results, vector<int> result, vector<int>& c, int target, int index){         for(int i = index; i < c.size(); ++i){             int t = target - c[i];             // if t is little than 0, then also the posibilities next are not necessary to check             if( t < 0){                 return;             }             else{                 // add c[i] to the result to check all the posibilities                 result.push_back(c[i]);                 if(t == 0)                     results.push_back(result);                 else                     helper(results, result, c, t, i);                 // pop c[i] to prepare for checking c[i+1]                 result.pop_back();             }//end of else         }//end of for     }//end of helper };

精简版

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class Solution { public:     vector<vector<int>> combinationSum(vector<int>& candidates, int target) {         vector<vector<int>> results;         sort(candidates.begin(), candidates.end());         helper(results,vector<int>{},candidates,target,0);         return results;     }     void helper(vector<vector<int>> &results, vector<int> result, vector<int>& c, int target, int index){         if(target == 0){             results.push_back(result);             return;         }         for(int i = index; i < c.size() && target >= c[i]; ++i){             result.push_back(c[i]);             helper(results, result, c, target - c[i], i);             result.pop_back();         }//end of for     }//end of helper };

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