0445. Add Two Numbers I I

445. Add Two Numbers II #

题目 #

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up: What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:


Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

题目大意 #

这道题是第 2 题的变种题,第 2 题中的 2 个数是从个位逆序排到高位,这样相加只用从头交到尾,进位一直进位即可。这道题目中强制要求不能把链表逆序。2 个数字是从高位排到低位的,这样进位是倒着来的。

解题思路 #

思路也不难,加法只用把短的链表依次加到长的链表上面来就可以了,最终判断一下最高位有没有进位,有进位再往前进一位。加法的过程可以用到递归。

代码 #


package leetcode

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers445(l1 *ListNode, l2 *ListNode) *ListNode {
	if l1 == nil {
		return l2
	}
	if l2 == nil {
		return l1
	}
	l1Length := getLength(l1)
	l2Length := getLength(l2)
	newHeader := &ListNode{Val: 1, Next: nil}
	if l1Length < l2Length {
		newHeader.Next = addNode(l2, l1, l2Length-l1Length)
	} else {
		newHeader.Next = addNode(l1, l2, l1Length-l2Length)
	}
	if newHeader.Next.Val > 9 {
		newHeader.Next.Val = newHeader.Next.Val % 10
		return newHeader
	}
	return newHeader.Next
}

func addNode(l1 *ListNode, l2 *ListNode, offset int) *ListNode {
	if l1 == nil {
		return nil
	}
	var (
		res, node *ListNode
	)
	if offset == 0 {
		res = &ListNode{Val: l1.Val + l2.Val, Next: nil}
		node = addNode(l1.Next, l2.Next, 0)
	} else {
		res = &ListNode{Val: l1.Val, Next: nil}
		node = addNode(l1.Next, l2, offset-1)
	}
	if node != nil && node.Val > 9 {
		res.Val++
		node.Val = node.Val % 10
	}
	res.Next = node
	return res
}

func getLength(l *ListNode) int {
	count := 0
	cur := l
	for cur != nil {
		count++
		cur = cur.Next
	}
	return count
}


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