Codewar记录一

问题描述：

 

 

 我的解决方案：

def get_sum(a,b):
    return sum(range(min(a, b), max(a, b) + 1))

 

一种巧妙的解决方案：

def get_sum(a, b):
    return (a + b) * (abs(a - b) + 1) // 2

 

 

第二个问题：

去除字符串的元音字母

法一：

def disemvowel(s):
    translator = str.maketrans({key: None for key in "aeiouAEIOU"})
    return s.translate(translator)

法二：

def disemvowel(s):
    for i in "aeiouAEIOU":
        s = s.replace(i,'')
    return s

法三：

def disemvowel(string):
    return "".join(c for c in string if c.lower() not in "aeiou")

 法四：

def disemvowel(s):
    translator = str.maketrans({key: None for key in "aeiouAEIOU"})
    return s.translate(translator)

 

 

问题三：求解连续子序列的最大值

def maxSequence(arr):
    max,curr=0,0
    for x in arr:
        curr+=x
        if curr<0:curr=0
        if curr>max:max=curr
    return max
max_sequence([-2, 1, -3, 4, -1, 2, 1, -5, 1000])

 

 

 

问题四：数字分解成质数幂的和

n = 12; decomp(12) -> "2^10 * 3^5 * 5^2 * 7 * 11"
since 12! is divisible by 2 ten times, by 3 five times, by 5 two times and by 7 and 11 only once.

n = 22; decomp(22) -> "2^19 * 3^9 * 5^4 * 7^3 * 11^2 * 13 * 17 * 19"

n = 25; decomp(25) -> 2^22 * 3^10 * 5^6 * 7^3 * 11^2 * 13 * 17 * 19 * 23

我的方案：

import math

def isPrime(n):

    if n <= 1:

        return False

    for i in range(2, int(math.sqrt(n)) + 1):

        if n % i == 0:

            return False

    return True


def decomp(n):
    #your code here
    m = math.factorial(n)
    str_  = ""
    for i in range(2,n+1):
        num = 0
        if isPrime(i):
            while (m%i == 0):
                m = m // i #这里注意是双分号，不然处理不了超大数字的除法
                num += 1
            if num == 1:
                str_ += "{0} * ".format(i)
            else:
                str_ += "{0}^{1} * ".format(i,num)
    str_ = str_.strip(" * ")
    return str_

decomp(3990)


其他人的解法：

from collections import defaultdict

def dec(n):
    decomp = defaultdict(lambda:0)
    i = 2
    while n > 1:
        while n % i == 0:
            n /= i
            decomp[i] += 1
        i += 1
    return decomp
            

def decomp(n):
    ans = defaultdict(lambda:0)
    for i in range(2, n + 1):
        for key, value in dec(i).items():
            ans[key] += value
    return ' * '.join('{}^{}'.format(x, y) if y > 1 else str(x) for x, y in sorted(ans.items()))

 

 问题5：

An anagram is the result of rearranging the letters of a word to produce a new word (see wikipedia).

Note: anagrams are case insensitive

Complete the function to return true if the two arguments given are anagrams of each other; return false otherwise.

Examples

• "foefet" is an anagram of "toffee"

• "Buckethead" is an anagram of "DeathCubeK"

#include <string>#先转换成集合，再对待比较的2个集合遍历集合中的每个元素，是狗相等，不等则提前退出
#include<algorithm>
using namespace std;
bool isAnagram(std::string test, std::string original){
  //your code here
  transform(original.begin(),original.end(),original.begin(),::tolower);
  transform(test.begin(),test.end(),test.begin(),::tolower);#大小写转换
  string s="";
  cout << original <<endl;
  for (unsigned long i=0;i<original.size();i++)
  {
      if(s.find(original[i],0)>10000)
      {
       s += original[i]; 
      }
  }
  for (unsigned long i=0;i<s.size();i++)
  {
    int num1 = 0;
    int num2 = 0;
    cout<<s[i]<<endl;
    for (unsigned long j=0;j<original.size();j++)
    {
      if(original[j]==s[i])
      {
        num1+=1;    
      }
    } 
    cout << num1 <<endl;
    for (unsigned long j=0;j<original.size();j++)
    {
      if(test[j]==s[i])
      {
        num2+=1;    
      } 
    }
    cout << num2 <<endl;
    if (num1!=num2) return false;
  }
    
  return true;
    
}

法二：

#include <string>

bool isAnagram(std::string t, std::string o){

 if(t.size() != o.size())
  return false;
  
  std::transform(t.begin(),t.end(),t.begin(),::tolower);
  std::transform(o.begin(),o.end(),o.begin(),::tolower);
  std::sort(t.begin(),t.end());
  std::sort(o.begin(),o.end());
  
  return t==o;
  
}

 

问题5：

 

 依次遍历：

//递归
def snail(snail_map):
    n = len(snail_map)
    new = []
    if (len(snail_map[0])==0):return new
    else:
        for i in range(n):
            new.append(snail_map[0][i])
        for i in range(1,n):
            new.append(snail_map[i][n-1])
        for i in range(n-2,-1,-1):
            new.append(snail_map[n-1][i])
        for i in range(n-2,0,-1):
            new.append(snail_map[i][0])    
        if (len(snail_map)-2<=0):return new
        else:
            map = []
            for i in range(1,n-1):
                map.append(snail_map[i][1:n-1])
            new.extend(snail(map))
    return new

//别人的方法;
def snail(array):
    ret = []
    if array and array[0]:
        size = len(array)
        for n in xrange((size + 1) // 2):
            for x in xrange(n, size - n):
                ret.append(array[n][x])
            for y in xrange(1 + n, size - n):
                ret.append(array[y][-1 - n])
            for x in xrange(2 + n, size - n + 1):
                ret.append(array[-1 - n][-x])
            for y in xrange(2 + n, size - n):
                ret.append(array[-y][n])
    return ret



//
def snail(array):
  if array:
    # force to list because zip returns a list of tuples
    top_row = list(array[0])
    # rotate the array by switching remaining rows & columns with zip
    # the * expands the remaining rows so they can be matched by column
    rotated_array = zip(*array[1:])
    # then reverse rows to make the formerly last column the next top row
    rotated_array = rotated_array[::-1]
    return top_row + snail(rotated_array)
  else:
    return []

问题6：数字转字符串

#include <string>

std::string number_to_string(int num) {
  std::string ret;
  bool neg = num < 0;
  num = abs(num);
  while(num) {
    ret += num % 10 + '0';
    num /= 10;
  }
  if(neg) ret += '-';
  std::reverse(ret.begin(), ret.end());
  return ret;
}

 

 问题7:2数之和问题，这个题在力扣见到过，就是给定一个数组和target，找到2数之和为target的对应的索引，如果有多个答案，则优先返回第一个。

C++是真的恶心，看书都能看懂，自己写代码全是错，要么就是左右的类型问题，比如for循环里，控制变量的值和初始化的值类型不同；要么就是const左右类型问题；要么就是没加域名限定符；然后各种STL的方法也没记得很清楚

 

C++返回元素对应的索引：

vector<double>::iterator ite1 = find(vec_dis.begin(), vec_dis.end(), vec_dis1[0]);
vector<double>::iterator ite2 = find(vec_dis.begin(), vec_dis.end(), vec_dis1[1]);
vector<double>::iterator ite3 = find(vec_dis.begin(), vec_dis.end(), vec_dis1[2]);
auto index1 = std::distance(std::begin(vec_dis), ite1);
auto index2 = std::distance(std::begin(vec_dis), ite2);
auto index3 = std::distance(std::begin(vec_dis), ite3); 

https://blog.csdn.net/sptoor/article/details/6615729常见的const迭代器有关内容

解决方案：

#include <vector>
#include <algorithm>
#include <iostream>


std::pair<std::size_t, std::size_t> two_sum(const//参数是const vector，所以后面的迭代器类型也应是const std::vector<int>& numbers, int target) {
  std::pair<std::size_t,std::size_t> p;
  std::vector<int> v{};
  for (unsigned long i=0;i<numbers.size();i++)  //迭代器的类型一般都是unsigned long
  {
    std::cout << target<<std::endl;
    if (std::find(v.cbegin(), v.cend(), target - numbers[i]) == v.cend())
    {
      v.push_back(numbers[i]);
    }    
    else
    {
      std::vector<int>::const_iterator ite1 = std::find(v.cbegin(), v.cend(),target - numbers[i]); 
      auto index1 = std::distance(std::cbegin(v), ite1);
      p.first = index1;
      p.second = i;
      return p;
    }
  }  
}

其他解决方案：

#include <unordered_map>

std::pair<std::size_t, std::size_t> two_sum(const std::vector<int>& numbers, int target) {
    std::unordered_map<int, std::size_t> counterparts;
    for (std::size_t i = 0; i < numbers.size(); ++i) {
        auto it = counterparts.find(numbers[i]);
        if (it != counterparts.end())
            return {it->second, i};
        counterparts.emplace(target - numbers[i], i);
    }
    return {-1, -1};
}