Question

所以，我正在编写一个代码来检查输入的字符串是否是回文，我已经编写了以下代码，但是它没有正常工作，例如，如果我输入＆＃34; race＆＃34;它仍然说是真的，虽然它应该说FALSE，请帮忙。这是代码

string = input("Please enter any word: ")
a = 0
string_length = len(string)
for string_index in range(string_length-1, -1, -1):
    character = string[string_index]
    if string[a] == character:
        a = a + 1
        b = "TRUE"
    else:
        b = "FALSE"
print(b)

Answer 1

正确的代码：

ClassCastException

Answer 2

除了this answer by מתן ל：

当反转时，您可以轻松地比较您的输入本身 - 列表理解使这一点变得微不足道：

简单的单字和案例感知测试：

library(shiny)
library(dplyr)

dat <- data.frame(outcome=sample(c("died","survived",NA), 20, TRUE),
                  cntr=sample(c("hospa","hospb"), 20, TRUE), 
                  s=rnorm(20), 
                  t=rnorm(20), stringsAsFactors=FALSE)

ui <- navbarPage(
  sidebarLayout(
    sidebarPanel(
      selectInput(inputId = "y", 
                  label = "Y-axis:",
                  choices = c("s"="s", "t"="t"), 
                  selected = "s"),
      selectInput(inputId = "z", 
                  label = "Color by:",
                  choices = c("outcome", "cntr"),
                  selected = "outcome")
    ),

    mainPanel(
      tabsetPanel(id="tabspanel", type = "tabs",
                  tabPanel(title = "Wilcox",
                  verbatimTextOutput(outputId = "p")))
    )
  )
)

server <- function(input, output, session) {
  df <- reactive({
    data.frame(dat[[input$y]], dat[[input$z]])
  })

  output$p <- renderPrint({
    wilcox.test(df()[,1] ~ df()[,2])
  })

}

shinyApp(ui=ui, server=server)

不区分大小写并允许标点符号

可从两侧读取回文。根据palindrom-ness的规则，你可能允许忽略套管，甚至腾出空白和标点符号。

将word = "SomemoS" print(word == word[::-1]) # word[::-1] simply reverses the word and prints True if same应用于
不关心空格和标点符号

lower()

Answer 3

string = input("Please enter any word: ")
string_length = len(string)
start_index = 0
count = 0 
end_index = string_length - 1
for string_index in range(int(string_length/2)):
    if string[start_index] == string[end_index]:
        start_index = start_index + 1
        end_index = end_index - 1
        count += 1
    else:
        pass
print("Palendrome") if count == int(string_length/2) else print("Not Palendrome")

说明：为了检查古院系统字符串，比较第一个和最后一个字符，直到比较到达字符串的中间

for string_index in range(int(string_length/2)):

比较第一个字符和最后一个字符：

if string[start_index] == string[end_index]:

更新start_index和end_index并计算：

start_index = start_index + 1
end_index = end_index - 1
count += 1

如果count等于字符串长度的一半那么它肯定是一个综合症

print("Palendrome") if count == int(string_length/2) else print("Not Palendrome")

输入是否是palidrome

3 个答案: