Question

def merge (seq, p, q, r):
    # n1: length of sub-array [p..q]
    n1 = q - p + 1
    # n2: length of sub-array [q+1 ..r]
    n2 = r - q
    # Left and Right arrays
    left_arr = [] 
    right_arr = []
    for i in xrange(0, n1):
        left_arr.append( seq[p+i] )
    for j in xrange(0, n2):
        right_arr.append( seq[q+j+1] )

    j=0
    i=0
    for k in xrange(p,r+1):
        if left_arr[i]<= right_arr[j]:
            seq[k]=left_arr[i]
            i+=1
        else:
            seq[k]=right_arr[j]
            j+=1
    return seq

s = [2,4,5,7,1,2,3,6]
p = 0
q = 3
r = 7
print merge(s,p,q,r)

工作原理：

取未排序的序列s，以及必须合并序列的索引号。（p =初始，r =最终，q =中）
现在，left_arr和right_arr分别是[p，q]，[q + 1，r]
我们采用left_arr和right_arr初始值（i = 0，j = 0）。我们迭代序列seq ...
迭代时我们用排序的值替换seq的值...

上面的函数能够排序到最后一个数字“7”..最后，它显示“IndexError”。我可以使用异常处理来捕获并修复但我认为...对于合并排序，它太多了..任何人都可以帮助我尽可能简单地修复代码。

我正在学习算法..（以下书：Thomas H. Cormen的算法导论）

Answer 1

问题是，在最后一次迭代中，我将等于4，并且您正在尝试访问不存在的left_arr [5]。当i或j变得大于相应数组的大小时，你应该添加一个停止条件，然后将另一个数组中的所有剩余元素添加到seq。

这是一个有效的代码：

def merge (seq, p, q, r):
    # n1: length of sub-array [p..q]
    n1 = q - p + 1
    # n2: length of sub-array [q+1 ..r]
    n2 = r - q
    # Left and Right arrays
    left_arr = seq[p:n1] #here 
    right_arr = seq[n1:r+1]  #here
    j=0
    i=0
    for k in xrange(p, r+1):
        if left_arr[i]<= right_arr[j]:
            seq[k]=left_arr[i]
            i+=1
            if i > n1-1: #here
                break
        else:
            seq[k]=right_arr[j] #here
            j+=1
            if j > n2-1:
                break
    if i >= len(left_arr): #from here down
        seq[k+1:] = right_arr[j:]
    elif j >= len(right_arr):
        seq[k+1:] = left_arr[i:]

    return seq

s = [2,4,5,7,1,1,1,1]
p = 0
q = 3
r = 7
print merge(s,p,q,r)

我已在评论中标记了我编辑代码的地方。

Answer 2

您的代码的问题在于您正在循环xrange(p, r+1)，因此在某个迭代中的循环期间，i或j的值可能会等于{的值{1}}或len(left)，最终导致len(right)。

IndexError

<强>输出：

def merge(seq,p,q,r):
    left=seq[p:q+1]       #a better way to fetch the list
    right=seq[q+1:]       
    i=0
    j=0
    #you shuldn't loop over the length of seq as that might make the value of either i or j
    # greater than the length of left or right lists respectively.
    # so you should only loop until one of the lists is fully exhausted

    while i<len(left) and j<len(right):
        if left[i] <= right[j] :
            seq[i+j]=left[i]
            i+=1
        else:
            seq[i+j]=right[j]
            j+=1

    #now the while loop is over so either right or left is fully traversed, which can be
    #find out my matching the value of j and i with lenghts of right and left lists respectively

    if j == len(right):
        seq[i+j:]=left[i:]  #if right is fully traversed then paste the remaining left list into seq
    elif i==len(left):      #this is just vice-versa of the above step
        seq[i+j:]=right[j:] 
    print seq

s = [2,4,5,7,1,2,3,6]
p = 0
q = 3
r = 7
merge(s,p,q,r)

Answer 3

在迭代left_arr和right_arr时，您没有检查数组索引。当另一个数组结束时，你应该合并任一数组的左边部分。

for k in xrange(p,r+1):
    if left_arr[i]<= right_arr[j]:
        seq[k]=left_arr[i]
        i+=1
    else:
        seq[k]=right_arr[j]
        j+=1

    # --------------- add this ----------------
    # if any array ends, merge the left elements of the other array
    if i >= len(left_arr):
        seq[k:] = right_arr[j:]
        break
    elif j >= len(right_arr):
        seq[k:] = left_arr[i:]
        break
    # --------------- end ----------------
return seq

合并排序索引错误

3 个答案: