python3字符串输出常见面试题总结

x=input("")
x=x[::-1] #列表切片，逆序输出
print(x)

n=int(input())
for i in range(n):
x=input()
print(x)
print()
while True:
x=input().split()
for i in x:
print(i)
print()

count = 1

while True:
  a, b = map(int, input().strip().split())
  # a和b都是0则退出
  if a == 0 and b == 0:
    break
  else:
    # 输入，把数据都存放在列表里，形成一个二维列表
    ls = []
    for i in range(a):
      ls1 = []
      n = input()
      for j in n:
        ls1.append(j)
      ls.append(ls1)
  # 对数据进行操作，先将所有的"."换成数字0，在判断周围有多少个雷，有就加1,判断周围的雷分别有4,5,8次
  for i in range(a):
    for j in range(b):
      if ls[i][j] == '.':
        ls[i][j] = 0
      if ls[i][j] == '*':
        # 判断左上角是否有雷
        if i - 1 >= 0 and j - 1 >= 0 and ls[i-1][j-1] != "*":
          ls[i-1][j-1] += 1
        # 判断左边是否有雷
        if j - 1 >= 0 and ls[i][j-1] != "*":
          ls[i][j-1] += 1
        # 判断左下角是否有雷
        if i + 1 <= a - 1 and j - 1 >= 0 and ls[i+1][j-1] != '*':
          ls[i+1][j-1] += 1
        # 判断上面是否有雷
        if i - 1 >= 0 and ls[i-1][j] != '*':
          ls[i-1][j] += 1
        # 判断下面是否有雷
        if i + 1 <= a-1 and ls[i+1][j] != '*':
          ls[i+1][j] += 1
        # 判断右上角是否有雷
        if i - 1 >= 0 and j + 1 <= b - 1 and ls[i-1][j+1] != '*':
          ls[i-1][j+1] += 1
        # 判断右边是否有雷
        if j + 1 <= b - 1 and ls[i][j+1] != '*':
          ls[i][j+1] += 1
        # 判断右下角是否有雷
        if i + 1 <= a - 1 and j + 1 <= b - 1 and ls[i+1][j+1] != "*":
          ls[i+1][j+1] += 1
  # 输出数据，并且统计执行次数
  print("Field #{}:".format(count))
  for i in ls:
    for j in i:
      print(j, end='')
    print()
  print()
  count += 1