1、二维数组的查找
查找,其实就可以挨个进行比较就可以。又由于题目说明(每一行都按照从左到右递增的顺序排序,每一列都按照从上到下递增的顺序排序),因此如果利用类似于二分查找的方法,那么比较次数则会更少。代码中以第一行最后一列的元素作为第一个比较的元素,比目标元素大则按行往左找,比目标元素小则按列往下找,直到找到或者下标溢出。
def Find(target, array): # write code here row = len(array) col = len(array[0]) i = 0 j = col - 1 while i < row and j >= 0: if target < array[i][j]: j = j - 1 elif target > array[i][j]: i = i + 1 else: return True return False array = [[1, 2, 8, 9], [2, 4, 9, 12], [4, 7, 10, 13], [6, 8, 11, 15]] target = 7 # row = len(array) # col = len(array[0]) # print(row, " ", col) print(Find(target, array))
2、替换空格
def replaceSpace(s): t = s.replace(' ', '%20') return t s = "We Are Happy" print(replaceSpace(s))
3、从头到尾打印链表
# -*- coding:utf-8 -*- class ListNode: def __init__(self, x): self.val = x self.next = None # 返回从尾部到头部的列表值序列,例如[1,2,3] def printListFromTailToHead(listNode): l = [] while listNode: l.append(listNode.val) listNode = listNode.next l.reverse() return l l1 = ListNode(1) l2 = ListNode(2) l3 = ListNode(3) l1.next = l2 l2.next = l3 l3.next = None l = printListFromTailToHead(l1) print(l)
4、重建二叉树
递归实现
# -*- coding:utf-8 -*- class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None class Solution: # 返回构造的TreeNode根节点 def reConstructBinaryTree(self, pre, tin): # write code here if len(pre) == 0: return None elif len(pre) == 1: return TreeNode(pre[0]) else: head = TreeNode(pre[0]) head.left = self.reConstructBinaryTree(pre[1:tin.index(pre[0]) + 1], tin[0:tin.index(pre[0])]) head.right = self.reConstructBinaryTree(pre[tin.index(pre[0]) + 1:], tin[tin.index(pre[0]) + 1:]) return head if __name__ == '__main__': s = Solution() pre = [1, 2, 4, 7, 3, 5, 6, 8] tin = [4, 7, 2, 1, 5, 3, 8, 6] ans = s.reConstructBinaryTree(pre, tin) print(ans.right.val)
5、用两个栈实现队列
栈:先进后出;队列:先进先出
用两个栈s1,s2实现队列,s1负责进队列,s2负责出队列。进队:进s1即可。出队:若s1,s2都为空,则队列为空;若s2不为空,则直接取s2最后一个元素;若s2为空,则把s1内元素按从尾到头依次放入s2,再取s2最后一个元素。
# -*- coding:utf-8 -*- class Solution: def __init__(self): self.Stack1 = [] self.Stack2 = [] def push(self, node): # write code here self.Stack1.append(node) def pop(self): # return xx if not self.Stack2 and not self.Stack1: return if self.Stack2: return self.Stack2.pop() else: while self.Stack1: self.Stack2.append(self.Stack1.pop()) return self.Stack2.pop() s = Solution() s.push(1) s.push(2) s.push(3) print(s.pop()) s.push(4) s.push(5) print(s.pop()) print(s.pop()) print(s.pop()) print(s.pop()) print(s.pop())