2. Add Two Numbers(两数相加)

解题方案

思路 1 ******- 时间复杂度: O(N)******- 空间复杂度: O(N)******

全部变成数字做加法再换回去呗,这多暴力,爽!

beats 54.35%

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if not l1:
            return l2
        if not l2:
            return l1
        
        val1, val2 = [l1.val], [l2.val]
        while l1.next:
            val1.append(l1.next.val)
            l1 = l1.next
        while l2.next:
            val2.append(l2.next.val)
            l2 = l2.next
            
        num1 = ''.join([str(i) for i in val1[::-1]])
        num2 = ''.join([str(i) for i in val2[::-1]])
        
        tmp = str(int(num1) + int(num2))[::-1]
        res = ListNode(int(tmp[0]))
        run_res = res
        for i in range(1, len(tmp)):
            run_res.next = ListNode(int(tmp[i]))
            run_res = run_res.next
        return res

思路 2 ******- 时间复杂度: O(N)******- 空间复杂度: O(1)******

可以使用递归,每次算一位的相加, beats 73.53%

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if not l1:
            return l2
        if not l2:
            return l1
        
        if l1.val + l2.val < 10:
            l3 = ListNode(l1.val+l2.val)
            l3.next = self.addTwoNumbers(l1.next, l2.next)
        else:
            l3 = ListNode(l1.val+l2.val-10)
            tmp = ListNode(1)
            tmp.next = None
            l3.next = self.addTwoNumbers(l1.next, self.addTwoNumbers(l2.next, tmp))
            
        return l3

来源:https://github.com/apachecn/awesome-algorithm/tree/master/docs/Leetcode_Solutions/Python

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转载自blog.csdn.net/mixiaolemy/article/details/84503608