算法导论学习笔记（1）

学习资源：网易公开课：算法导论

http://open.163.com/special/opencourse/algorithms.html

运行时间：输入本身、输入size、want 运行时间上界

分析方法：1、worst-case analysis(usually): T(n):maximium time on any input of size n

                                 Depends on computer:-------relative speed(same machine)/absolute speed(different)

                  2、average-case(sometime):T(n):期望时间

                  3、best-case analysis(bogus（假象）)

渐进分析(asymptotic analysis):忽略机器常量影响、关注增长as n->∞

渐进符号

theta符号θ：放弃低阶项、忽略常数因子：3n3+90n2-5n+81=θ(n3)

插入排序  T（n）=nΣ(j=2)   θ(j)=θ (n2)

merge sort归并排序：递归树

示例程序(python):

import datetime

a=[25,54,56,987,8,5,2,6,9,13,32,34,76,3,1,9898,23,234,555543,75,4]
b=[25,54,56,987,8,5,2,6,9,13,32,34,76,3,1,9898,23,234,555543,78,4]
i=1
starttime = datetime.datetime.now()
while i<len(a):
    j=i
    while j-1>=0:
        #print(j-1,j)
        if a[j-1]>a[j]:
            tmp=a[j-1]
            a[j-1]=a[j]
            a[j]=tmp
        j-=1
    i+=1
#output

endtime = datetime.datetime.now()
print("insert sort cost time:",endtime - starttime)
print(a)
    
def merge(seq):
    if len(seq)==1:
        #print(seq)
        return seq
    seq1=merge(seq[0:int(len(seq)/2)])
    seq2=merge(seq[int(len(seq)/2):])
    ll=list()
    i=j=0
    while True:
        if i==len(seq1) and j==len(seq2):
            return ll
        elif i==len(seq1):
            ll.append(seq2[j])
            j+=1
        elif j==len(seq2):
            ll.append(seq1[i])
            i+=1
        elif seq1[i]<seq2[j]:
            ll.append(seq1[i])
            i+=1
        elif seq2[j]<seq1[i]:
            ll.append(seq2[j])
            j+=1
starttime = datetime.datetime.now()
bb=merge(b)

endtime = datetime.datetime.now()
print("merge sort cost time:",endtime - starttime)
print(bb)

running example:

insert sort cost time: 0:00:00
[1, 2, 3, 4, 5, 6, 8, 9, 13, 23, 25, 32, 34, 54, 56, 75, 76, 234, 987, 9898, 555543]
merge sort cost time: 0:00:00
[1, 2, 3, 4, 5, 6, 8, 9, 13, 23, 25, 32, 34, 54, 56, 76, 78, 234, 987, 9898, 555543]
>>>