算法题目1

# greedy algorithm

money = [100,50,20,5,1]

def change_money(x):
    change = [0,0,0,0,0]
    for i,m in enumerate(money):
        change[i] = x // money[i]
        x = x % money[i]
    if x > 0:
        print("还剩%s" % x)
    return change

print(change_money(356.2))

def fib(n):
    ‘‘‘斐波那契数列‘‘‘
    f = [1,1]
    for i in range(2, n+1):
        f.append(f[-1]+f[-2])
    print(f)
    return f[n]

fib(5)

def LIS(x):
    ‘‘‘最长公共子序列‘‘‘
    F = [0 for _ in range(len(x))]
    p = [-1 for _ in range(len(x))]
    # 初始化
    F[0] = 1
    p[0] = -1
    for k in range(1, len(F)):
        max_loc = -1
        max_num = 0
        # 内层循环表示F[0:k]里所有小于x[k]的对应位置的F[i]的最大值
        for i in range(0, k):
            if x[i] < x[k]:
                if F[i] > max_num:
                    max_loc = i
                    max_num = F[i]
        F[k] = max_num + 1
        p[k] = max_loc

    max_i = 0
    for i in range(1,len(F)):
        if F[i] > F[max_i]:
            max_i = i

    lis = []
    i = max_i
    while i >= 0:
        lis.append(x[i])
        i = p[i]
    lis.reverse()
    return lis

print(LIS([9,7,2,8,3,5,2]))

def LCS(x, y):
    # 最长公共子序列（二维）
    F = [[0 for _ in range(len(y)+1)] for _ in range(len(x)+1)]
    p = [[0 for _ in range(len(y)+1)] for _ in range(len(x)+1)]
    for i in range(1, len(x)+1):
        p[i][0] = 2
    for j in range(1, len(y)+1):
        p[0][j] = 1

    # 0 斜向  1 横向 j-1   2竖向 i-1
    for i in range(1, len(x)+1):
        for j in range(1, len(y)+1):
            if x[i-1] == y[j-1]:
                F[i][j] = F[i-1][j-1]+1
                p[i][j] = 0
            else:
                #F[i][j] = max(F[i-1][j], F[i][j-1])
                if F[i-1][j] > F[i][j-1]:
                    F[i][j] = F[i-1][j]
                    p[i][j] = 2
                else:
                    F[i][j] = F[i][j-1]
                    p[i][j] = 1

    lcs = []
    i = len(x)
    j = len(y)
    while i > 0 or j > 0:
        if p[i][j] == 0:
            lcs.append(x[i-1])
            i -= 1
            j -= 1
        elif p[i][j] == 1:
            j -= 1
        else:
            i -= 1
    lcs.reverse()
    return lcs
    #return F[i][j]

print(LCS("ABBCBDE", "DBBCDB"))

def LCSS_BF(x,y):
    # 最长公共字符串（暴力）
    m = len(x)
    n = len(y)
    max_len = 0
    max_str = ""
    for k in range(0, min(m,n)):
        for i in range(0, m-k+1):
            for j in range(0, n-k+1):
                if x[i:i+k] == y[j:j+k]:
                    if k > max_len:
                        max_len = k
                        max_str = x[i:i+k]
    return max_str

print(LCSS_BF("ABBCBDE", "DBBCDB"))

def LCSS(x, y):
    # 最长公共字符串
    F = [[0 for _ in range(len(y)+1)] for _ in range(len(x)+1)]
    p = [[0 for _ in range(len(y)+1)] for _ in range(len(x)+1)]
    # 0 不匹配 1匹配
    for i in range(1, len(x)+1):
        for j in range(1, len(y)+1):
            if x[i-1] == y[j-1]:
                F[i][j] = F[i-1][j-1]+1
                p[i][j] = 1
            else:
                F[i][j] = 0
                p[i][j] = 0
    max_val = 0
    max_i = 0
    max_j = 0
    for i in range(1, len(x)+1):
        for j in range(1, len(y)+1):
            if F[i][j] > max_val:
                max_val = F[i][j]
                max_i = i
                max_j = j
    #tracback
    lcss = []
    i = max_i
    j = max_j
    while p[i][j] == 1:
        lcss.append(x[i-1])
        i -= 1
        j -= 1

    lcss.reverse()
    return lcss

print(LCSS("ABBCBDE", "DBBCDB"))