算法练习题

1.TwoSum

  给一个int型数组,要求找出其中两个和为特定值得数的坐标

  注意点:

    返回的左表一要比坐标二小

    最小的坐标是1,不是0

  例子:

  输入:numbers={2,7,11,15},target=9 输出:index1=1,index2=2

解:

def twoSum(nums,target):
    hash_map = {}
    for index,value in enumerate(nums):
        hash_map[value] = index
    print(hash_map)
    for index1,value in enumerate(nums):
        if target - value in hash_map:
            index2 = hash_map[target-value]
            if index1 != index2:
                return [index1+1,index2+1]
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2.Add Two Numbers

  定义这样的一个链表,链表的每个节点都存有一个0-9的数字,把链表当成数字,表头为高位,表尾为低位。如1->2->3表示321,现在要对两个这样的链表求和。

  注意点:

    数字的高低位,应该从低位向高位进位

    有多种情况要考虑,如链表长度是否相等、是否进位等

  例子:

  输入:(2->4->3)+(5->6->4) 输出:7->0->8

解:

class ListNode:
    def __init__(self,x):
        self.val = x
        self.next = None
        
    def myPrint(self):
        print(self.val)
        if self.next:
            self.next.myPrint()


def addTwoNodes(n1,n2):
    if not n1 and not n2:
        pass
    if not n1:
        return n2.val
    if not n2:
        return n1.val
    return n1.val+n2.val

def addTwoNumbers(l1,l2):
    result = ListNode(0)
    cur = result
    while l1 or l2:
        cur.val += addTwoNodes(l1,l2)
        if cur.val >=10:
            cur.val -= 10
            cur.next = ListNode(1)
        else:
            if l1 and l1.next or l2 and l2.next:
                cur.next = ListNode(0)
        cur = cur.next
        if l1:
            l1 = l1.next
        if l2:
            l2 = l2.next
    return result

lst = ListNode(9)
lst.next = ListNode(8)
addTwoNumbers(lst,ListNode(1)).myPrint()
View Code
原文地址:https://www.cnblogs.com/muchengQ/p/11939362.html