数据结构应用题答案第7章10

def display(self):
for i in range(self.len):
print(self.list[i].key,end=' ')
print()
def merge(self,order,a,i,k,j):
order[i] = a[i]
def mergeSort(self):
s = 1 # 已排序的子序列的长度，初始值为1
order = [ None ] * self.len
while s<self.len: # 归并过程
t = i
m = i
n = k + 1
while m<=k and n<=j:
# 将具有较小关键字值的元素放入order[]
if a[m].key<=a[n].key:
p = p+2*s
if p+s-1<n-1: # 归并长度不等的有序表
self.merge(order,a,p,p+s-1,n-1)
else: # 将一个有序表中的元素放入order[]中
for i in range(p,n):
def insert(self,i,x):
# 在第i个位置之前插入记录x
if self.len==self.maxSize:
raise Exception("顺序表已满")
if i<0 or i>self.len:
t += 1
n += 1
def mergepass(self,order,a,s,n):
p = 0
while p+2*s-1<=n-1: # 两两归并长度均为s的有序表
self.merge(order,a,p,p+s-1,p+2*s-1)
sl = SqList(50)
data = [1,2,5,4,7,6,3,0]
for i,x in zip(range(len(data)),data):
sl.insert(i,RecordNode(x,x))
sl.mergeSort()
sl.display()
class RecordNode(object):
def __init__(self,key,data):
self.key = key # 关键字
self.data = data # 数据元素的值
class SqList(obБайду номын сангаасect):
n += 1
while m<=k:
order[t] = a[m]
t += 1
m += 1
while n<=j:
order[t] = a[n]
order[t] = a[m]
t += 1
m += 1
else:
order[t] = a[n]
t += 1
self.mergepass(order,self.list,s,self.len)
s = s * 2
self.mergepass(self.list,order,s,self.len)
s = s * 2
raise Exception("插入位置不合理")
for j in range(self.len,i,-1):
self.list[j] = self.list[j-1]
self.list[i] = x
self.len += 1
def __init__(self,maxSize):
self.maxSize = maxSize # 顺序表的最大存储空间
self.list = [ None ] * self.maxSize # 待排序的记录集合
self.len = 0 # 顺序表的长度