python算法之查找

Posted 2020-11-25 xhw19950606

#顺序查找
#基本思想：从第一个元素到最后一个元素依次查找
def sqsearch(numList, x):
    for id,num in enumerate(numList):
        if num == x:
            return id
    return str(x) + ‘ is not exist!‘
print(sqsearch([1,2,4,5,6,8,12,24,32,44], 5))

#二分查找（折半查找）
#基本思想：将n个数组成的有序数列分成个数大致相同的两半，取a[n/2]与欲查找的x作比较，
#         如果x=a[n/2]则找到x，算法终止；如果x<a[n/2]，则只要在数组a的左半部继续搜索x；
#         如果x>a[n/2]，则只要在数组a的右半部继续搜索x
def binarysearch(numList, x):
    low = 0
    high = len(numList) - 1
    while low <= high:
        mid = (low + high) / 2
        if x == numList[mid]:
            return mid
        if x > numList[mid]:
            low =mid + 1
        if x < numList[mid]:
            high = mid - 1
    return str(x) + " is not exist!"
print(binarysearch([1,2,4,5,6,8,12,24,32,44], 5))

#插入查找（按比例查找）
#基本思想：属于二分查找的改进方法，不同之处是将折半的取法变为按比例
def insertsearch(numList, x):
    low = 0
    high = len(numList) - 1
    while low <= high:
        mid = low + (x-numList[low])/(numList[high]-numList[low])*(high-low)#插入查找公式
        if x == numList[mid]:
            return mid
        if x > numList[mid]:
            low = mid + 1
        if x < numList[mid]:
            high = mid - 1
    return str(x) + " is not exist!"
print(insertsearch([1, 2, 4, 5, 6, 8, 12, 24, 32, 44], 5))

#斐波那契查找（黄金比例查找）
#基本思想：属于二分查找的改进方法，不同之处是将折半的取法变为0.618:1的取法（以列表的元素个数用斐波那契数列来近似）
def fib(x):
    if x==1 or x==2:
        return 1
    else:
        return fib(x-1) + fib(x-2)
def find(key):
    x=1
    fibnum = []
    while fib(x)<key:
        fibnum.append(fib(x))
        x +=1
    fibnum.append(fib(x))
    return fibnum
def fibsearch(numList, x):
    org=orgleng = len(numList)
    #若原始列表长度不足斐波那契数，则用列表最后一个数补，直到长度等于斐波那契数（注：只需要补1次就行）
    fibnum = find(orgleng)
    while orgleng < fibnum[-1]:
        numList.append(numList[-1])
        orgleng += 1
    k = len(fibnum)
    low = 0
    high = len(numList) - 1
    while low <= high:
        mid = low + fib(k-1)-1#按照斐波那契数列取值划分
        print(str(numList[low:mid]) + ‘ ‘ + str(numList[mid:high+1]))
        if x > numList[mid]:
            low = mid + 1
            k -= 2 #当x处于后半段时
        elif x < numList[mid]:
            high = mid - 1
            k -= 1 #当x处于前半段时
        else:
            #若mid小于原始列表最大索引，则返回；若mid大于原始列表最大索引，则返回原始列表最大索引
            if mid < org:
                return mid
            else:
                return org - 1
    return str(x) + " is not exist!"
print(fibsearch([1, 2, 4, 5, 6, 8, 12, 24, 32, 44],2))
# [1, 2, 4, 5, 6, 8, 12] [24, 32, 44, 44, 44, 44]
# [1, 2, 4, 5] [6, 8, 12]
# [1, 2] [4, 5]
# [1] [2]