两个单链表的求和,将矩阵元素全部对换
发布日期:2021-06-29 11:42:20 浏览次数:2 分类:技术文章

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def lengthNum(head):    length = 0    pt = head    while pt:        length += 1        pt = pt.next    return lengthdef computeSum(pt1, pt2, count):    if pt1 == None or pt2 == None:        return 0    if count > 0:        flag = computeSum(pt1.next, pt2, count - 1)        result = pt1.val + flag    else:        flag = computeSum(pt1.next, pt2.next, 0)        result = pt1.val + pt2.val + flag    pt1.val = result % 10    return result // 10def sumOf2Num(head1, head2):    if head1 == None:        return head2    if head2 == None:        return head1    length1 = lengthNum(head1)    length2 = lengthNum(head2)    if length1 < length2:        head1, head2 = head2, head1        length1, length2 = length2, length1    pt1 = head1    pt2 = head2    flag = computeSum(pt1, pt2, length1 - length2)    if flag == 0:        return head1    else:        new_node = ListNode(1)        new_node.next = head1        return new_node
def rotate(matrix):    cols = len(matrix)    rows = len(matrix[0])        for i in range(cols):        for j in range(rows - 1, i - 1, -1):            matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]                for i in range(cols):        matrix[i].reverse()            return matrix

 

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感谢大佬
[***.8.128.20]2024年04月04日 06时11分45秒