49. Group Anagrams
Medium
Given an array of strings, group anagrams together.
Example:
Input:["eat", "tea", "tan", "ate", "nat", "bat"]
, Output: [ ["ate","eat","tea"], ["nat","tan"], ["bat"] ]
bonus:
1.join function usage:
str = "-"; seq = ("a", "b", "c"); # 字符串序列 print str.join( seq );
result:
a-b-c
class Solution: def groupAnagrams(self, strs): res = {} for item in strs: k = ''.join(sorted(item)) # 字符串排序 print("k") print(k) print("sorted") print(sorted(item)) if k not in res: # 判断是否存在 res[k] = [] res[k].append(item) # 相同字符串放到同一个字典 k中 return [res[x] for x in res] # 输出结果 if __name__ == '__main__': strs = ['eat', 'tea', 'tan', 'ate', 'nat', 'bat'] solu = Solution() print(solu.groupAnagrams(strs))
C++ version Analyse:
This time appears some small problem such as forget iterator,new variable during two-dimension vector loop forget to define.
#include<stdio.h> #include<iostream> #include<string> #include<vector> #include<set> #include<map> #include<algorithm> using namespace std; class Solution { public: vector<vector<string>> groupAnagrams(vector<string>& strs) { //map get value map<string,vector<string>> temp; vector<vector<string>> res; //middle handle for(int i=0;i<strs.size();i++) { string str=strs[i]; sort(str.begin(),str.end()); if(temp.find(str)==temp.end()) { vector<string> tmp1={}; temp[str]=tmp1; } temp[str].push_back(strs[i]); // cout<<str<<endl; } //Assign the second value of the map to the vector. for(map<string,vector<string>>::iterator it=temp.begin();it!=temp.end();it++) { //pushback is a function it need a variable to get the value it output by.push_back res.push_back((*it).second); } return res; } }; int main() { Solution s; vector<string> strs={"eat", "tea", "tan", "ate", "nat", "bat"}; vector<vector<string>> res; res=s.groupAnagrams(strs); //cout<<res<<endl; //output result for(vector<vector<string>>::iterator ite=res.begin();ite!=res.end();ite++) { //whenever use a new variable remember to define variables vector<string> tmp_vec=*ite; //iterator variable must add ::iterator for(vector<string>::iterator it=tmp_vec.begin();it!=tmp_vec.end();it++) { cout<<*it<<" "; } cout<<endl; } return 0; }
Special Attention the two-dimensional loop way,please don't check next time:
for(vector<vector<string>>::iterator ite=res.begin();ite!=res.end();ite++) { //whenever use a new variable remember to define variables vector<string> tmp_vec=*ite; //iterator variable must add ::iterator for(vector<string>::iterator it=tmp_vec.begin();it!=tmp_vec.end();it++) { cout<<*it<<" "; } cout<<endl; }
50. Pow(x, n)
Medium
Implement pow(x, n), which calculates x raised to the power n (xn).
Example 1:
Input: 2.00000, 10 Output: 1024.00000
Example 2:
Input: 2.10000, 3 Output: 9.26100
Example 3:
Input: 2.00000, -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25
Note:
- -100.0 < x < 100.0
n is a 32-bit signed integer, within the range [−231, 231 − 1]
Method Idea:
Use recursion to finish this problem logn,divide the solution into first juge the minus or positive,then only if index=0 return 1,other situation multiply x or divide into small parts
#include<stdio.h> #include<iostream> #include<string> #include<vector> #include<set> #include<algorithm> using namespace std; class Solution { public: double myPow(double x, int n) { return n>=0?myPowImp(x,n):1.0/myPowImp(x,n);} private: double myPowImp(double x,int n){ if(n==0) return 1; return myPowImp(x*x,n/2)*(n%2?x:1); } }; int main() { Solution s; double x=2.00000; int n=10; double res=s.myPow(x,n); cout<<res<<endl; return 0; }
Python version has the same idea:
Bonus:
1.recursive itself should add self.myPow(x,n-1) use self to add the function name
2.three special version,if minus,let minus to x-->1/x,change n-->-n
3.judge whether n%2 to x*x,n/2 or (x,n-1)*x
class Solution(object): def myPow(self, x, n): if n==0: return 1 if n<0: x=1/x n=-n if n%2: return x*self.myPow(x,n-1) return self.myPow(x*x,n/2) solu=Solution() x=2 n=10 print(solu.myPow(x,n))
for(vector<vector<string>>::iterator ite=res.begin();ite!=res.end();ite++) { //whenever use a new variable remember to define variables vector<string> tmp_vec=*ite; //iterator variable must add ::iterator for(vector<string>::iterator it=tmp_vec.begin();it!=tmp_vec.end();it++) { cout<<*it<<" "; } cout<<endl; }
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