11 二进制中1的个数
输入一个整数,输出该数二进制表示中1的个数。其中负数用补码表示。
# -*- coding:utf-8 -*-
import sys
class Solution:
def NumberOf1(self, n):
# write code here
count=0
n=int(n)
if n<0:
n= n & 0xffffffff
while n:
n=(n-1)& n
count+=1
return count
12 数值的整数次方
给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。
# -*- coding:utf-8 -*-
class Solution:
def Power(self, base, exponent):
# write code here
return base**exponent
13 调整数组顺序使奇数位于偶数前面
输入一个整数数组,实现一个函数来调整该数组中数字的顺序,使得所有的奇数位于数组的前半部分,所有的偶数位于数组的后半部分,并保证奇数和奇数,偶数和偶数之间的相对位置不变。
# -*- coding:utf-8 -*-
class Solution:
def reOrderArray(self, array):
# write code here
arr=array
n=len(arr)
ord,even=[],[]
#n=len(arr)
for i in range(n):
if arr[i]%2==1:
ord.append(arr[i])
else:
even.append(arr[i])
return ord+even
14 链表中倒数第k个结点
输入一个链表,输出该链表中倒数第k个结点。
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def FindKthToTail(self, head, k):
# write code here
l=[]
current=head
while current!=None:
l.append(current)
current=current.next
if k>len(l) or k<1:
return
return l[-k]
15 反转链表
输入一个链表,反转链表后,输出新链表的表头。
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# 返回ListNode
def ReverseList(self, pHead):
# write code here
if not pHead or not pHead.next:
return pHead
last=None
while pHead:
temp=pHead.next
pHead.next=last
last=pHead
pHead=temp
return last
16 合并两个排序的链表
输入两个单调递增的链表,输出两个链表合成后的链表,当然我们需要合成后的链表满足单调不减规则。
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# 返回合并后列表
def Merge(self, pHead1, pHead2):
# write code here
mergeHead=ListNode(1)
p=mergeHead
while pHead1 and pHead2:
if pHead1.val >=pHead2.val:
mergeHead.next=pHead2
pHead2=pHead2.next
else:
mergeHead.next=pHead1
pHead1=pHead1.next
mergeHead=mergeHead.next
if pHead1:
mergeHead.next=pHead1
elif pHead2:
mergeHead.next=pHead2
return p.next
17 树的子结构
输入两棵二叉树A,B,判断B是不是A的子结构。(ps:我们约定空树不是任意一个树的子结构)
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def HasSubtree(self, pRoot1, pRoot2):
# write code here
if not pRoot1 or not pRoot2 :
return False
return self.is_subtree(pRoot1,pRoot2) or self.HasSubtree(pRoot1.left,pRoot2) or self.HasSubtree(pRoot1.right,pRoot2)
def is_subtree(self,pRoot1,pRoot2):
if not pRoot2:
return True
if not pRoot1 or pRoot1.val!= pRoot2.val:
return False
return self.is_subtree(pRoot1.left,pRoot2.left) and self.is_subtree(pRoot1.right,pR
18 二叉树的镜像
操作给定的二叉树,将其变换为源二叉树的镜像。
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回镜像树的根节点
def Mirror(self, root):
# write code here
if not root:
return None
else:
root.left,root.right=root.right,root.left
self.Mirror(root.left)
self.Mirror(root.right)
19 顺时针打印矩阵
输入一个矩阵,按照从外向里以顺时针的顺序依次打印出每一个数字,例如,如果输入如下4 X 4矩阵: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 则依次打印出数字1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10.
# -*- coding:utf-8 -*-
class Solution:
# matrix类型为二维列表,需要返回列表
def printMatrix(self, matrix):
# write code here
res=[]
n=len(matrix) #行数
m=len(matrix[0]) #列数
if n==1 and m==1:
res=[matrix[0][0]]
return res
for i in range(min(n,m)+1//2):
[res.append(matrix[i][j]) for j in range(i,m-i) if matrix[i][j] not in res]
[res.append(matrix[k][m-1-i]) for k in range(i,n-i) if matrix[k][m-1-i] not in res]
[res.append(matrix[n-i-1][l]) for l in range(m-i-1,i-1,-1) if matrix[n-i-1][l] not in res]
[res.append(matrix[q][i]) for q in range(n-i-1,i-1,-1)if matrix[q][i] not in res]
return res
20 包含min函数的栈
定义栈的数据结构,请在该类型中实现一个能够得到栈中所含最小元素的min函数(时间复杂度应为O(1))。
# -*- coding:utf-8 -*-
class Solution:
def __init__(self):
self.stack=[]
self.min_stack=[]
def push(self, node):
# write code here
self.stack.append(node)
if not self.min_stack or node <= self.min_stack[-1]:
self.min_stack.append(node)
def pop(self):
# write code here
if self.min_stack[-1]== self.stack[-1]:
self.min_stack.pop()
self.stack.pop()
def top(self):
# write code here
return self.stack[-1]
def min(self):
# write code here
return self.min_stack[-1]