python3的各种经典案例，总共299个案例，直接可以运行（前：100个案例）
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一. python3的各种经典案例，总共299个案例，直接可以运行（前：100个案例）
二. python3的各种经典案例，总共299个案例，直接可以运行（中：100个案例）
三. python3的各种经典案例，总共299个案例，直接可以运行（后：99个案例）
第一章 难度等级★
【例1】反转一个3位整数 难度等级★
3.代码实现
class Solution:
#参数number: 一个三位整数
#返回值: 反转后的数字
def reverseInteger(self, number):
h = int(number/100)
t = int(number%100/10)
z = int(number%10)
return (100*z+10*t+h)
#主函数
if __name__ == '__main__':
solution = Solution()
num = 123
ans = solution.reverseInteger(num)
print("输入：", num)
print("输出：", ans)
【例2】合并排序数组 难度等级★
3.代码实现
class Solution:
#参数A: 有序整数数组A
#参数B: 有序整数数组B
#返回:一个新的有序整数数组
def mergeSortedArray(self, A, B):
i, j = 0, 0
C = []
while i < len(A) and j < len(B):
if A[i] < B[j]:
C.append(A[i])
i += 1
else:
C.append(B[j])
j += 1
while i < len(A):
C.append(A[i])
i += 1
while j < len(B):
C.append(B[j])
j += 1
return C
#主函数
#主函数
if __name__ == '__main__':
A = [1,4]
B = [1,2,3]
D = [1,2,3,4]
E = [2,4,5,6]
solution = Solution()
print("输入：", A, " ", B)
print("输出：", solution.mergeSortedArray(A,B))
print("输入：", D, " ", E)
print("输出：", solution.mergeSortedArray(D,E))
【例3】旋转字符串 难度等级★
3.代码实现
class Solution:
#参数s:字符列表
#参数offset:整数
#返回值:无
def rotateString(self, s, offset):
if len(s) > 0:
offset = offset % len(s)
temp = (s + s)[len(s) - offset : 2 * len(s) - offset]
for i in range(len(temp)):
s[i] = temp[i]
#主函数
if __name__ == '__main__':
s = ["a","b","c","d","e","f","g"]
offset = 3
solution = Solution()
solution.rotateString(s, offset)
print("输入：s =", ["a","b","c","d","e","f","g"], " ", "offset =",offset)
print("输出：s =", s)
【例4】相对排名 难度等级★
3.代码实现
class Solution:
#参数nums为整数列表
#返回列表
def findRelativeRanks(self, nums):
score = {}
for i in range(len(nums)):
score[nums[i]] = i
sortedScore = sorted(nums, reverse=True)
answer = [0] * len(nums)
for i in range(len(sortedScore)):
res = str(i + 1)
if i == 0:
res = 'Gold Medal'
if i == 1:
res = 'Silver Medal'
if i == 2:
res = 'Bronze Medal'
answer[score[sortedScore[i]]] = res
return answer
#主函数
if __name__ == '__main__':
num = [5,4,3,2,1]
s = Solution()
print("输入为：",num)
print("输出为：",s.findRelativeRanks(num))
【例5】二分查找 难度等级★
3.代码实现
class Solution:
#参数nums: 整数数组
#参数target: 要查找的目标数字
#返回值：目标数字的第一个位置，从0开始
def binarySearch(self, nums, target):
return self.search(nums, 0, len(nums) - 1, target)
def search(self, nums, start, end, target):
if start > end:
return -1
mid = (start + end)//2
if nums[mid] > target:
return self.search(nums, start, mid, target)
if nums[mid] == target:
return mid
if nums[mid] < target:
return self.search(nums, mid, end, target)
#主函数
if __name__ == '__main__':
my_solution = Solution()
nums = [1,2,3,4,5,6]
target = 3
targetIndex = my_solution.binarySearch(nums, target)
print("输入：nums =", nums, " ", "target =",target)
print("输出：",targetIndex)
【例6】下一个更大的数 难度等级★
3.代码实现
class Solution:
#参数nums1为整数数组
#参数nums2为整数数组
#返回整数数组
def nextGreaterElement(self, nums1, nums2):
answer = {}
stack = []
for x in nums2:
while stack and stack[-1] < x:
answer[stack[-1]] = x
del stack[-1]
stack.append(x)
for x in stack:
answer[x] = -1
return [answer[x] for x in nums1]
#主函数
if __name__ == '__main__':
s = Solution()
nums1 = [4,1,2]
nums2 = [1,3,4,2]
print("输入1为：",nums1)
print("输入2为：",nums2)
print("输出为 ：",s.nextGreaterElement(nums1,nums2))
【例7】字符串中的单词数 难度等级★
3.代码实现
class Solution:
#参数s为字符串
#返回整数
def countSegments(self, s):
res = 0
for i in range(len(s)):
if s[i] != ' ' and (i == 0 or s[i - 1] == ' '):
res += 1
return res
#主函数
if __name__ == '__main__':
s = Solution()
n = "Hello, my name is John"
print("输入为：",n)
print("输出为：",s.countSegments(n))
【例8】勒索信 难度等级★
3.代码实现
class Solution:
"""
参数ransomNote为字符串
参数magazine为字符串
返回布尔类型
"""
def canConstruct(self, ransomNote, magazine):
arr = [0] * 26
for c in magazine:
arr[ord(c) - ord('a')] += 1
for c in ransomNote:
arr[ord(c) - ord('a')] -= 1
if arr[ord(c) - ord('a')] < 0:
return False
return True
#主函数
if __name__ == '__main__':
s = Solution()
ransomNote = "aa"
magazine = "aab"
print("输入勒索信为：",ransomNote)
print("输入杂志内容：",magazine)
print("输出为：",s.canConstruct(ransomNote,magazine))
【例9】不重复的两个数 难度等级★
3.代码实现
#参数arr是输入的待查数组
#返回值是两个值的列表，内容没有重复的
class Solution:
def theTwoNumbers(self, a):
ans = [0, 0]
for i in a:
ans[0] = ans[0] ^ i
c = 1
while c & ans[0] != c:
c = c << 1
for i in a:
if i & c == c:
ans[1] = ans[1] ^ i
ans[0] = ans[0] ^ ans[1]
return ans
if __name__ == '__main__':
arr = [1, 2, 5, 1]
solution = Solution()
print(" 数组为：", arr)
print(" 两个没有重复的数字是:", solution.theTwoNumbers(arr))
【例10】双胞胎字符串 难度等级★
3.代码实现
#参数s和t是一对字符串
#返回值是个字符串，能否根据规则转换
class Solution:
def isTwin(self, s, t):
if len(s) != len(t):
return "No"
oddS = []
evenS = []
oddT = []
evenT = []
for i in range(len(s)):
if i & 1:
oddS.append(s[i])
oddT.append(t[i])
else :
evenS.append(s[i])
evenT.append(t[i])
oddS.sort()
oddT.sort()
evenS.sort()
evenT.sort()
for i in range (len(oddS)) :
if oddS[i] != oddT[i]:
return "No"
for i in range (len(evenS)) :
if evenS[i] != evenT[i]:
return "No"
return "Yes"
if __name__ == '__main__':
s="abcd"
t="cdab"
solution = Solution()
print(" s与t分别为：", s, t)
print(" 是否:", solution.isTwin(s, t))
【例11】最接近target的值 难度等级★
3.代码实现
#参数array是输入列表
#参数target是目标值
#返回值是整数
class Solution:
def closestTargetValue(self, target, array):
n = len(array)
if n < 2:
return -1
array.sort()
diff = 0x7fffffff
left = 0
right = n - 1
while left < right:
if array[left] + array[right] > target:
right -= 1
else:
diff = min(diff, target - array[left] - array[right])
left += 1
if diff == 0x7fffffff:
return -1
else:
return target - diff
if __name__ == '__main__':
array = [1,3,5,11,7]
target = 15
solution = Solution()
print(" 输入数组为：", array,"目标值为：", target)
print(" 最近可以得到值为:", solution.closestTargetValue(target, array))
【例12】点积 难度等级★
3.代码实现
#参数A和B是输入列表
#返回值是个整数，是点积
class Solution:
def dotProduct(self, A, B):
if len(A) == 0 or len(B) == 0 or len(A) != len(B):
return -1
ans = 0
for i in range(len(A)):
ans += A[i] * B[i]
return ans
if __name__ == '__main__':
A = [1,1,1]
B = [2,2,2]
solution = Solution()
print(" A与B分别为：", A, B)
print(" 点积为：", solution.dotProduct(A, B))
【例13】函数运行时间 难度等级★
3.代码实现
#参数s为输入原始字符串
#返回值是个字符串，意为每个名字的函数运行了多久
class Solution:
def getRuntime(self, a):
map={}
for i in a:
count = 0
while not i[count] == ' ':
count = count + 1
fun = i[0 : count]
if i[count+2] == 'n':
count = count + 7
v = int(i[count:len(i)])
if fun in map.keys():
map[fun] = v - map[fun]
else:
map[fun] = v
else:
count = count + 6
v = int(i[count:len(i)])
map[fun] = v - map[fun]
res=[]
for i in map:
res.append(i)
res.sort()
for i in range(0,len(res以上是关于python3的各种经典案例，总共299个案例，直接可以运行（前：100个案例）的主要内容，如果未能解决你的问题，请参考以下文章
python3的各种经典案例，总共299个案例，直接可以运行（中：100个案例）
python3的各种经典案例，总共299个案例，直接可以运行（前：100个案例）
python3的各种经典案例，总共299个案例，直接可以运行（后：99个案例）
python3的各种经典案例，总共299个案例，直接可以运行（中：100个案例）