宝宝刷 leetcode

12/3

1、Two Sum

Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dict = {}
        for i in range(len(nums)):
            j = target - nums[i]
            if j in dict:
                return [dict[j],i]
            else:
                dict[nums[i]] = i 
        #return 0

2. Add Two Numbers

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        ans = 0
        unit = 1
        while l1 or l2:
            if l1:
                ans += l1.val * unit
                l1 = l1.next
            if l2:
                ans += l2.val * unit
                l2 = l2.next
            unit *= 10

        alpha = cur = ListNode(0)

        for n in reversed(str(ans)):
            cur.next = ListNode(int(n))
            cur = cur.next
    
        return alpha.next

补充：链表是由一些节点构成的，这些节点之间由指针连接，形成了一个链式结构。最基本的链表节点只需要存储当前节点的值，和一个指向下一节点的指针。由这种只存储下一节点地址的链表节点构成的链表被称为单向链表。

在节点ListNode定义中，定义为节点为结构变量。节点存储了两个变量：value 和 next。value 是这个节点的值，next 是指向下一节点的指针，当 next 为空指针时，这个节点是链表的最后一个节点。构造函数包含两个参数 _value 和 _next ，分别用来给节点赋值和指定下一节点。

struct ListNode {
       int val;    //定义val变量值，存储节点值
       struct ListNode *next;   //定义next指针，指向下一个节点，维持节点连接
  }

203. Remove Linked List Elements

Input:  1->2->6->3->4->5->6, val = 6
Output: 1->2->3->4->5

class Solution:
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        if head == None:
            return head
        dummy = ListNode(0)
        dummy.next = head
        pre = dummy
        while head:
            if head.val == val:
                pre.next = head.next
                head = pre
            pre = head
            head = head.next
        return dummy.next

206. Reverse Linked List

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

class Solution:
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        tail = None
        cur = head
        while cur:
            next_node = cur.next
            cur.next = tail
            tail = cur
            cur = next_node
        return tail

26. Remove Duplicates from Sorted Array

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

class Solution:
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        cnt=0
        if (len(nums)==0):
            return cnt 

        for i in sorted(set(nums)):
            nums[cnt]=i
            cnt+=1
        
        return cnt

27. Remove Element

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

class Solution:
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        index = 0
        while(index < len(nums)):
            if (nums[index] == val):
                nums.pop(nums.index(val))
            else:
                index += 1

88. Merge Sorted Array

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3

Output: [1,2,2,3,5,6]

class Solution:
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: void Do not return anything, modify nums1 in-place instead.
        """
        for v in nums2:
            nums1[m] = v
            m+=1
        nums1.sort()

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