Combination Sum(leetcode)
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</pre><pre code_snippet_id="516969" snippet_file_name="blog_20141112_1_5676001" name="code" class="cpp">
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]
问题描述如上,简单说就是找出数组中a1,a2....an中能够组合成目标和的个数,每一个数可以取多次。
采用深度优先搜索实现,先将数组排序,从最小的元素开始
如图所示:
图中红色为剪枝的节点,不会保留下来。
有人可能会问,纹身模蓝色节点5处,为什么是从3开始,因为上一次选择的是3,所以不会出现下次的选择小于上次的,这样就不会出现 2,3,2之类的
也就是说选择出来的结果也是有序的
代码如下:
<span style="font-size:18px;">class Solution {public: vector<vector<int> > combinationSum(vector<int> &candidates, int target) { sort(candidates.begin(),candidates.end()); vector<vector<int>> result; vector<int> temp; int i; dfs(candidates,target,0,temp,result); return result; } void dfs(vector<int> &candidates,int target,int start,vector<int> &temp,vector<vector<int>> &result) { if(target==0) { result.push_back(temp); return; } for(int i=start;i<candidates.size();i++) { if(target<candidates[i]) return; temp.push_back(candidates[i]); dfs(candidates,target-candidates[i],i,temp,result); temp.pop_back(); } }};</span>
如果,数组的元素只能够选择一次呢,即相同的元素只能够选择一次
那么我们需要将当前选择的元素保存下来,下次选择时先判断是否当前元素等于保留的元素,如果是,则跳过,否则继续
代码如下:
<span style="font-size:18px;">class Solution {public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { sort(num.begin(),num.end()); vector<int> temp; vector<vector<int>> result; dfs(num,target,0,temp,result); return result; } void dfs(vector<int>&num,int target,int start,vector<int> &temp,vector<vector<int>> &result) { if(target==0) { result.push_back(temp); return; } int remark=0; for(int i=start;i<num.size();i++) { if(target<num[i]) return; if(remark==num[i]) continue; remark=num[i]; temp.push_back(num[i]); dfs(num,target-num[i],i+1,temp,result); temp.pop_back(); } }};</span>
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