【LeetCode】2. Add Two Numbers

LeetCode2 傳送門

解法：

    初始化一個dummy_head，一個新的鏈表curr用於保存結果，逐位遍歷兩鏈表節點，將計算結果保存到curr鏈表中，直到兩鏈表節點均爲空。遍歷結束後carry非0，則將curr的next指向一個新的值爲carry的鏈表節點。

Python源碼：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        dummy_head = ListNode(0)
        p = l1
        q = l2
        curr = dummy_head
        carry = 0
        while p or q:
            if p:
                x = p.val
            else:
                x = 0
            if q:
                y = q.val
            else:
                y = 0
            digit = carry + x + y
            carry = digit // 10
            curr.next = ListNode(digit % 10)
            curr = curr.next
            if p:
                p = p.next
            if q:
                q = q.next
        if carry > 0:
            curr.next = ListNode(carry)
        return dummy_head.next

我的心路：

Runtime: 64 ms, faster than 22.34% of Python online submissions for Add Two Numbers.

Memory Usage: 12 MB, less than 13.97% of Python online submissions for Add Two Numbers.

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        c = False
        res = l1
        while l1 != None and l2 != None:
            l1.val += l2.val
            if c:
                l1.val += 1
                c = False
            if l1.val >= 10:
                l1.val %= 10
                c = True
            cur = l1
            l1 = l1.next
            l2 = l2.next
        
        if l2 != None:
            l1 = l2
            cur.next = l1
            
        while l1 != None:
            if c:
                l1.val += 1
                c = False
            if l1.val >= 10:
                l1.val %= 10
                c = True
            cur = l1
            l1 = l1.next
        
        if c:
            cur.next = ListNode(1)
        
        return res

    七個月前的解法，看起來比現在可讀性高一些，加入了dummy head，全面碾壓這次的解法：

Runtime: 60 ms, faster than 41.16% of Python online submissions for Add Two Numbers.

Memory Usage: 11.8 MB, less than 61.76% of Python online submissions for Add Two Numbers.

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        l3 = l = ListNode(0)
        car = 0
        while(l1 and l2):
            l.next = ListNode((l1.val + l2.val + car) % 10)
            car = (l1.val + l2.val + car) // 10
            l1 = l1.next
            l2 = l2.next
            l = l.next
            
        while(l1):
            l.next = ListNode((l1.val + car) % 10)
            car = (l1.val + car) // 10
            l1 = l1.next
            l = l.next
        while(l2):
            l.next = ListNode((l2.val + car) % 10)
            car = (l2.val + car) // 10
            l2 = l2.next
            l = l.next
        if(car):
            l.next = ListNode(car)
            
        return l3.next