代码随想录刷题随记23-回溯3-编程知识

代码随想录刷题随记23-回溯3

39. 组合总和

leetcode链接
注意同一个数字可以无限制重复被选取
怎么体现这个可以重复取的思想很重要
解题代码：

class Solution {
public:void backtrace( vector<vector<int>>& ret,vector<int> &path,vector<int>& candidates,int target,int index,int &sum){if(sum==target){ret.push_back(path);return ;}if(sum>target)return;for(int i=index;i<candidates.size();i++) {  path.push_back(candidates[i]);sum+=candidates[i];//重用backtrace(ret, path,candidates,target, i, sum);sum-=candidates[i];path.pop_back();}}vector<vector<int>> combinationSum(vector<int>& candidates, int target) {vector<vector<int>> ret;vector<int> path;int sum=0;backtrace(ret,path, candidates, target,  0, sum);return ret;}
};

40.组合总和II

leetcode链接
candidates 中的每个数字在每个组合中只能使用一次
同时因为candidite集合里面的数本来就有一样的，所以有肯能虽然取数顺寻不一致，但是结果集合相同的问题：
在这里插入图片描述

需要去重
去重的关键在于同一层不能重复，纵向上可以重复：
在这里插入图片描述
可以先对candidates进行排序，这样比较容易跳过
解题代码：

class Solution {
public:void backtrace( vector<vector<int>>& ret,vector<int>& candidates,int target,int sum,vector<int> & path,int index,vector<bool> &use){if(target==sum){ret.push_back(path);return;  }if(sum>target)return;for(int i=index;i<candidates.size();i++){  if(i>0&&candidates[i]==candidates[i-1]&&(use[i-1]==false)) continue;  path.push_back(candidates[i]);sum+=candidates[i];use[i]=true;backtrace(ret,candidates,  target,  sum, path, i+1,use);sum-=candidates[i];use[i]=false;path.pop_back();}        }vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {vector<vector<int>> ret;vector<int> path;int sum=0;std::sort(candidates.begin(),candidates.end());vector<bool> use(candidates.size(),false);backtrace(ret, candidates, target, sum,path, 0,use);return ret;}
};

131.分割回文串

leetcode链接

class Solution {
public:bool isstr(string str){if(str.size()==1)return true;int l=0;int r=str.size()-1;while(l<r){if(str[l]!=str[r])return false;l++;r--;}return true;}void backtrace( string s,vector<vector<string>>& ret,vector<string> &path,int curind,int lastind){if(curind==s.size()){ret.push_back(path);return;}string substring=s.substr(lastind,curind-lastind);//从当前切substring+=s[curind];if(isstr(substring)){path.push_back(substring);backtrace(s, ret, path,  curind+1, curind+1);path.pop_back();}substring.pop_back();//当前不切if(curind+1<s.size())backtrace(s, ret, path,  curind+1, lastind);}vector<vector<string>> partition(string s) {vector<vector<string>> ret;vector<string> path;backtrace(s, ret, path, 0, 0);return ret;}
};