Python每日一练(20230420) 数组逐位判断、交错字符串、二进制求和

# -*- 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)

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 = ""))

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"))