***(leetcode_backtracking) Combination Sum

Combination Sum

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Question Solution 

Given a set of candidate numbers (C) 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.
• 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 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

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 Array Backtracking

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元素可重复，那就开始 用target/candidates[i]， 然后将可能需要的candidates[i] 添加到vector中。

和Combination Sum II 没什么区别

class Solution {    vector<bool> flag;    vector<vector<int> > ret;    void dfs(vector<int> &candidates, int target, int cursum, int curDep, int maxDep){        if(cursum==target){            vector<int> ans;            for(int i=0;i<maxDep;i++)                if(flag[i])                    ans.push_back(candidates[i]);            ret.push_back(ans);            return;        }        if(curDep>=maxDep||cursum>target)            return;        if((curDep==0)||(curDep!=0 &&(((candidates[curDep]==candidates[curDep-1])&&flag[curDep-1])||candidates[curDep]!=candidates[curDep-1]))){            flag[curDep]=true;            dfs(candidates, target, cursum+candidates[curDep], curDep+1, maxDep);        }        flag[curDep]=false;        dfs(candidates, target, cursum, curDep+1, maxDep);    }public:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        int len = candidates.size();        for(int i=0;i<len;i++){            int num = target/candidates[i];            for(int j=1;j<num;j++)                candidates.push_back(candidates[i]);        }        for(int i=0;i<candidates.size();i++)            flag.push_back(false);        sort(candidates.begin(),candidates.end());        dfs(candidates, target, 0, 0, candidates.size());        return ret;            }};