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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]
先排序!
这个题,去重是关键;//[2,2,2,2] 4 怎么快速去重呢?
先抛出个很笨重的去重吧。40ms
解法1:
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(),candidates.end());
vector<int> temp;
vector<vector<int>> res;
combinationSum(candidates,0,target,temp,res);
return res;
}
private:
void combinationSum(vector<int>& candidates,int index,int target,vector<int> temp,vector<vector<int>> &res){
if(index==candidates.size()||candidates[index]>target) return;//终止条件
if(candidates[index]==target){
temp.push_back(candidates[index]);
if(res.size()>0){//skip duplicates too slow!!!
for(int i=0;i<res.size();i++){
if(equal(temp,res[i])) return;
}
}//不重复就加入!//但是这判断重复的代价太大了吧???
res.push_back(temp);//找到一组满足条件的
return;
}
temp.push_back(candidates[index]);
combinationSum(candidates,index+1,target-candidates[index],temp,res);
temp.pop_back();
combinationSum(candidates,index+1,target,temp,res);
}
bool equal(vector<int> &a,vector<int> &b){
if(a.size()!=b.size()) return false;
for(int i=0;i<a.size();i++){
if(a[i]!=b[i]) return false;
}
return true;
}
};
每次都是把整个res中的vector全部比较一遍,时间成本很高。
要想出在刚开始就去掉的方法!
解法2:把后面相同的元素跳过去。20ms
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(),candidates.end());
vector<int> temp;
vector<vector<int>> res;
combinationSum(candidates,0,target,temp,res);
return res;
}
private:
void combinationSum(vector<int>& candidates,int index,int target,vector<int> temp,vector<vector<int>> &res){
if(index==candidates.size()||candidates[index]>target) return;//终止条件
for(int i=index;i<candidates.size();i++){
if(i>index&&candidates[i]==candidates[i-1])
continue;
temp.push_back(candidates[i]);
if(candidates[i]==target){
res.push_back(temp);//找到一组满足条件的
return;
}
combinationSum(candidates,i+1,target-candidates[i],temp,res);
temp.pop_back();
}
}