Question

您好我是python的新手，我试图通过在python中编写线性搜索来学习它，这里是代码，这是我到目前为止编写的代码。

a = [100]

n = int(raw_input("Enter the no. of elements: "))

print "Enter the elements: "
for i in range(0,n):
    print i

item = int(raw_input("Enter the item to be searched: "))

for i in range(0,10):

    if a[i] == item:
        print "Item ", item, "found at location", i+1
        break;

if i == n:
    print "Item", item, "not found"

我收到一个错误，说列表索引超出范围，我哪里错了？

Answer 1

列表中有一个元素。第二个外观尝试访问超出范围的元素1到9。

我建议使用enumerate，它返回（index，element）对，直到迭代的长度

for i, x in enumerate(a):
    if x == item:
        # found item at position i

如果要输入数字以生成元素列表，则

n = int(raw_input("Enter the no. of elements: "))
# build your list of n elements here 
for i in range(0,n):
    print i

例如，a = range(0, n)或导入random模块，如果您不想要线性数字范围

Answer 2

您的代码中存在一些错误，我尝试重新排列它：

a = []

n = int(raw_input("Enter the no. of elements: "))

print "Enter the elements: "
for i in range(0,n):
    a.append(i)
    print (i)

item = int(raw_input("Enter the item to be searched: "))

found = False
for i in range(0,len(a)):
    if a[i] == item:
        found = True
        print "Item ", item, "found at location", i+1
        break;

if (not found):
    print "Item", item, "not found"

请注意以下几点：

使用列表而不是数组：a = []
使用列表操作：例如a.append(i)将元素插入第一个空闲插槽中的列表
使用布尔值
发现所有python功能;）

Answer 3

略有不同（排序算法）

a = []

n = int(input("enter number of elements: "))

print("appending elements: ")

for i in range(0, n):
    a.append(i)
    print(i)

item = int(input("enter the item to be searched: "))
# for python2.x please use raw_input

for i in range(len(a)):
    print("searching...")
    if a[i] == item:
        print("found! at: %s" % i)
        break
    else:
        print("pretty bad day, nothing found!")

使用python进行线性搜索

3 个答案: