Python每日一练(20230420)

目录

1. 数组逐位判断  

2. 交错字符串  

3. 二进制求和  

4. 四舍六入五成双规则  

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1. 数组逐位判断

比如有以下数组:
a1: 1,0,0,1,0,0,0,1
a2: 0,0,0,0,1,1,1,1
a3: 0,1,0,1,0,1,0,0
a4: 1,0,1,1,1,1,0,0
a5: .......

抓取三个数组进行判断, if ((a1第一位or a2第一位 or a3第一位=1 )and (a1第二位 or a2 第二位 or a3第二位=1)and.... 直到判断完所有位数为止,所有位都有了1的话就输出当前这三个数组,已输出的数组不参与之后的判断。

出处:

https://edu.csdn.net/practice/26046536

代码:

# -*- coding: UTF-8 -*-
from itertools import combinations
a1=[ 1,0,0,1,0,0,0,1]
a2=[ 0,0,0,0,1,1,1,1]
a3=[ 0,1,0,1,0,1,0,0]
a4=[ 1,0,1,1,1,1,0,0]
a5=[ 1,1,1,1,1,1,1,0]
a6=[ 0,0,0,0,0,0,0,1]
a=[a1,a2,a3,a4,a5,a6]
al = list(combinations(a,3))
for i in al:
    flag = True
    for j in range(len(i[0])):
        if (i[0][j] + i[1][j] + i[2][j] == 0):
            flag = False
            break
    if flag:
        print(i)

输出:

([1, 0, 0, 1, 0, 0, 0, 1], [0, 0, 0, 0, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 0])
([1, 0, 0, 1, 0, 0, 0, 1], [0, 1, 0, 1, 0, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0])
([1, 0, 0, 1, 0, 0, 0, 1], [1, 0, 1, 1, 1, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0])
([1, 0, 0, 1, 0, 0, 0, 1], [1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1])
([0, 0, 0, 0, 1, 1, 1, 1], [0, 1, 0, 1, 0, 1, 0, 0], [1, 0, 1, 1, 1, 1, 0, 0])
([0, 0, 0, 0, 1, 1, 1, 1], [0, 1, 0, 1, 0, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0])
([0, 0, 0, 0, 1, 1, 1, 1], [1, 0, 1, 1, 1, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0])
([0, 0, 0, 0, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1])
([0, 1, 0, 1, 0, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1])
([1, 0, 1, 1, 1, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1])


2. 交错字符串

给定三个字符串 s1s2s3,请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。

两个字符串 s 和 t 交错 的定义与过程如下,其中每个字符串都会被分割成若干 非空 子字符串:

  • s = s1 + s2 + ... + sn
  • t = t1 + t2 + ... + tm
  • |n - m| <= 1
  • 交错 是 s1 + t1 + s2 + t2 + s3 + t3 + ... 或者 t1 + s1 + t2 + s2 + t3 + s3 + ...

提示:a + b 意味着字符串 a 和 b 连接。

示例 1:

输入:s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出:true

示例 2:

输入:s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出:false

示例 3:

输入:s1 = "", s2 = "", s3 = ""
输出:true

提示:

  • 0 <= s1.length, s2.length <= 100
  • 0 <= s3.length <= 200
  • s1s2、和 s3 都由小写英文字母组成

出处:

https://edu.csdn.net/practice/26046537

代码:

class Solution(object):
    def isInterleave(self, s1, s2, s3):
        """
        :type s1: str
        :type s2: str
        :type s3: str
        :rtype: bool
        """
        if len(s1) + len(s2) != len(s3):
            return False
        queue = [(0, 0), (-1, -1)]
        visited = set()
        isSuccess = False
        index = 0
        while len(queue) != 1 or queue[0][0] != -1:
            p = queue.pop(0)
            if p[0] == len(s1) and p[1] == len(s2):
                return True
            if p[0] == -1:
                queue.append(p)
                index += 1
                continue
            if p in visited:
                continue
            visited.add(p)
            if p[0] < len(s1):
                if s1[p[0]] == s3[index]:
                    queue.append((p[0] + 1, p[1]))
            if p[1] < len(s2):
                if s2[p[1]] == s3[index]:
                    queue.append((p[0], p[1] + 1))
        return False
# %%
s = Solution()
print(s.isInterleave(s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"))
print(s.isInterleave(s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"))
print(s.isInterleave(s1 = "", s2 = "", s3 = ""))

输出:

True
False
True


3. 二进制求和

给你两个二进制字符串,返回它们的和(用二进制表示)。

输入为 非空 字符串且只包含数字 1 和 0

示例 1:

输入: a = "11", b = "1"
输出: "100"

示例 2:

输入: a = "1010", b = "1011"
输出: "10101"

提示:

  • 每个字符串仅由字符 '0' 或 '1' 组成。
  • 1 <= a.length, b.length <= 10^4
  • 字符串如果不是 "0" ,就都不含前导零。

出处:

https://edu.csdn.net/practice/26046539

代码:

class Solution(object):
    def addBinary(self, a, b):
        res = ''
        lsa, lsb = len(a), len(b)
        pos, plus, curr = -1, 0, 0
        while (lsa + pos) >= 0 or (lsb + pos) >= 0:
            if (lsa + pos) >= 0:
                curr += int(a[pos])
            if (lsb + pos) >= 0:
                curr += int(b[pos])
            res = str(curr % 2) + res
            curr //= 2
            pos -= 1
        if curr == 1:
            res = '1' + res
        return res
# %%
s = Solution()
print(s.addBinary(a = "11", b = "1"))
print(s.addBinary(a = "1010", b = "1011"))

输出:

100
10101


4. 四舍六入五成双规则

输入:1234
输出:1234
12 12.0
4 4.00
0.2 0.200
0.32 0.320
1.3 1.30
1.235 1.24
1.245 1.24
1.2451 1.25

出处:

https://edu.csdn.net/practice/26046539

代码:

nums = [1234,12,4,0.2,0.32,1.3,1.235,1.245,1.2451]
for n in nums:
    a = str(n)
    if '.' in a:
        a = float(a)
        if a*1000%10!=5:
            a = '%.2f'%(a)
        else:
            if len(str(a).split(".")[1])>3:
                a = '%.2f'%(a)
            else:
                if int(a*100%10%2)==1:
                    a = float('%.2f'%float(int(a*100)/100))+0.01
                else:
                    a = '%.2f'%float(int(a*100)/100)
        print(a)
    else:
        a = int(a)
        if a>99:
            print(a)
        else:
            if 0 < a < 10: 
                print('%.2f'%a)
            else:
                print(float(a))

输出:

1234
12.0
4.00
0.20
0.32
1.30
1.24
1.24
1.25


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