【LeetCode】39. Combination Sum & 40. Combination Sum II分析及解法&DFS
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39. Combination Sum
Total Accepted: 87334 Total Submissions: 283947 Difficulty: Medium
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]
【题意】
从给定的数据集合中寻找合适的若干数据(可重复利用),使得这些数据之和等于目标值(target)
【分析】
很显然,解此题需要用到深度优先搜索(Deepth First Search,DFS ),关于DFS的定义百度可以查到:点击打开链接,本题中,由于数据可重复利用,我们需要注意循环中的搜索起点的设置。当搜索进行到数据集末端(类似二叉树搜索中的叶子节点),我们需要采用“回溯法”,返回一层继续搜索。
【解法】
class Solution{public:vector<vector<int>> combinationSum(vector<int>& nums, int target){vector<int> temp;//存储可能满足要求的搜索路径sort(nums.begin(),nums.end());//首先将数组进行升序排序inputNumSet=nums;//复制输入数据集合DFS(temp,target, 0);//从第一个元素开始深度优先搜索return res;}private:vector<vector<int>> res;//存储所有满足要求的搜索路径,即满足要求的数据组合vector<int> inputNumSet;//保存输入数据集//计算当前搜索路径数据之和int sumOfNums(vector<int>& temp){int sum=0;for(int i=0;i<temp.size();i++){sum+=temp[i];}return sum;}//深度优先搜索void DFS(vector<int>& temp, int target, int start){int sum;sum=sumOfNums(temp);//计算目前搜索路径数据之和if(sum==target){res.push_back(temp);//如果目前组合数据和与目标相等,则存入容器resreturn;}else if(sum>target){return;//如果目前组合已经大于目标值,则对此路径进行“剪枝”,直接返回,不再继续此路径}else{//如果当前组合数值比目标值小,则继续搜索for(int i=start;i<inputNumSet.size();i++){if(i!=start&&inputNumSet[i]==inputNumSet[i-1])//避免重复解,由于给定的数据集中可能存在相同数据,continue;temp.push_back(inputNumSet[i]);DFS(temp,target,i);//由于数据可重复利用,继续搜索时,起始点依旧为i,如果不允许重复利用数据,这里应为i+1temp.pop_back();//上面一句为递归调用,要么找到目标路径并存储而返回,要么路径不满足要求而返回。 //返回之后,进行回溯,返回继续搜索}}}};
40. Combination Sum II
Total Accepted: 65829 Total Submissions: 239855 Difficulty: MediumGiven 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]
【题意】
从给定的数据集合中寻找合适的若干数据(不可重复利用),使得这些数据之和等于目标值(target),然后返回所有可能的搜索路径。
【分析】
此题和上面一题非常类似,解答方法基本一致,这里不再赘述,关键注意:如何避免重复解?
【解法】
class Solution{public:vector<vector<int>> combinationSum2(vector<int>& nums, int target){vector<int> temp;//存放搜索路径sort(nums.begin(),nums.end());//升序排序inputNumSet=nums;//存放输入数据DFS(temp,target,0);//进行深度优先搜索,求取目标路径并存入res容器return res;}private://【数据域】vector<vector<int>> res;vector<int> inputNumSet;//【成员函数域】//函数1:求取当前数据组合的和int sumOfNums(vector<int>& temp){int sum=0;for(int i=0;i<temp.size();i++)sum+=temp[i];return sum;}//函数2:深度优先搜索,寻找和与目标值相等数据组合void DFS(vector<int>& temp,int target,int start){//if(start==inputNumSet.size())return;//搜索到叶子节点,返回int sum;sum=sumOfNums(temp);if(sum==target)//搜索到目标路径,将当前路径存入结果容器,返回{res.push_back(temp);return;}else if(sum>target)//当前路径结果超过目标值,返回{return;}else//当前路径结果小于目标值,则继续搜索{for(int i=start;i<inputNumSet.size();i++){if(i!=start&&inputNumSet[i]==inputNumSet[i-1])//避免重复组合continue;temp.push_back(inputNumSet[i]);DFS(temp,target,i+1);//递归调用,要么找到目标路径并存储而返回,要么路径不满足要求而返回temp.pop_back();//返回之后,进行回溯,返回继续搜索}}}};
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