lintCode 入门题 python版解答

献给我的人生第二春，下面开始问题与答案

矩阵面积

class Rectangle():
'''
* Define a constructor which expects two parameters width and height here.
'''
width = 0.1
height= 0.1
# write your code here
def __init__(self, width,height):
self.width = width
self.height = height
'''
* Define a public method `getArea` which can calculate the area of the
* rectangle and return.
'''
# write your code here
def getArea(self):
return self.width*self.height

2.二叉树遍历

二叉树原理点击打开链接

其它解答http://www.cnblogs.com/bozhou/p/LintCode.html

　　LintCode已实现NodeTree类，传入的参数实际为一个NodeTree类型的二叉树，没有找到方法接触源码，只能通过网页一步一步调试，分析出结构：
其中，Node节点类拥有三个属性
　　left：当前节点的左节点
　　right：当前节点的右节点
　　val：当前节点的值

class Solution:
"""
@param: root: the root of tree
@return: the max node
"""
maxVal = -9999
node = None
# write your code here
def maxNode(self, root):
if root == None:
return None
self.max(root)
return self.node

def max(self,node):
if node == None:
return None
if node.val > self.maxVal:
self.node = node
self.maxVal = node.val
self.max(node.left)
self.max(node.right)

3 ,整数排序

class Solution:
"""
@param: A: an integer array
@return:
"""
def sortIntegers(self, A):
# write your code here
for i in range(len(A)-1):
for j in range(len(A)-i-1):
if A[j]>A[j+1]:
B=A[j]
A[j]=A[j+1]
A[j+1]=B
return A