Leetcode--Combination Sum II

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.
• 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 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

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 Array Backtracking

Have you met this question in a real interview? 

思路：还是沿着问题1的解题思路，需要额外注意的是重复问题，如：[1,1,1] target=1,得到的结果不能是三个[1]，那么如果消除重复呢，最简单的方法就是用set容器

class Solution {public:    vector<vector<int>> res;    set<vector<int>> result;    void backtrace(vector<int> v,vector<int>&num,int target,int pos,int sum)    {        if(sum==target)            result.insert(v);//res.push_back(v);        else if(sum>target||pos>=num.size())            return ;        else if(sum<target){            for(int i=pos;i<num.size();i++)            {                vector<int> tmp(v);                tmp.push_back(num[i]);                backtrace(tmp,num,target,i+1,sum+num[i]);            }        }    }    vector<vector<int> > combinationSum2(vector<int> &num, int target) {        if(num.size()<1)            return res;                    sort(num.begin(),num.end());        vector<int> v;        backtrace(v,num,target,0,0);        res.assign(result.begin(),result.end());        return res;    }};Submission Result: Time Limit ExceededLast executed input:[6,23,31,29,16,25,14,7,14,17,9,24,24,13,26,14,30,26,24,16,20,8,27,25,16,14,18,12,7,27,16,9,7,13,25,19,33,26,21,5,9,18,16,20,26,7,21,21,21,32,24,28,27,17,18,9,18,23,16,20,15,19,16,9], 24

但是得到TLE，这是怎么回事呢？

分析可能的原因在于backtrace中连续声明并初始化tmp对象，带来了额外的时间开销。

优化该段代码：

class Solution {public:    vector<vector<int>> res;    set<vector<int>> result;    void backtrace(vector<int> &v,vector<int>&num,int target,int pos,int sum)    {        if(sum==target)            result.insert(v);//res.push_back(v);        else if(sum>target||pos>=num.size())            return ;        else if(sum<target){            for(int i=pos;i<num.size();i++)            {                                v.push_back(num[i]);                backtrace(v,num,target,i+1,sum+num[i]);                v.pop_back();            }        }    }    vector<vector<int> > combinationSum2(vector<int> &num, int target) {        if(num.size()<1)            return res;                    sort(num.begin(),num.end());        vector<int> v;        backtrace(v,num,target,0,0);        res.assign(result.begin(),result.end());        return res;    }};Submission Result: Accepted