LeetCode开心刷题二十六天——49.Group Anagrams50. Pow(x, n)

49. Group Anagrams

Medium

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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

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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;
    }