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go语言刷题笔记---- 两数相加

干自闭 2022-02-27 阅读 37

文章目录

原题

给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。

请你将两个数相加,并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外,这两个数都不会以 0 开头。

提示:

每个链表中的节点数在范围 [1, 100] 内
0 <= Node.val <= 9
题目数据保证列表表示的数字不含前导零

解题

第一次尝试方法:
将两个链表的数据转换成两个整数,然后相加得到sum,再将sum拆分成链表返回

package main

import "fmt"

type ListNode struct {
	Val  int
	Next *ListNode
}

/**
*	头插入数据 A --->  B->A ---> C->B->A
 */
func insert(head *ListNode, val int) *ListNode {
	newList := new(ListNode)
	newList.Val = val
	newList.Next = head
	return newList
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
	//遍历链表 l1 获取num1
	num1 := 0
	var index int = 1
	for l1 != nil {
		num1 += l1.Val * index
		l1 = l1.Next
		index *= 10
	}
	fmt.Printf("num1: %v\n", num1)
	//遍历链表 l2 获取num2
	num2 := 0
	var index2 int = 1
	for l2 != nil {
		num2 += l2.Val * index2
		l2 = l2.Next
		index2 *= 10
	}
	fmt.Printf("num2: %v\n", num2)
	//两书相加
	num := num1 + num2
	fmt.Printf("num: %v\n", num)
	//申请头结点
	sumList := new(ListNode)
	sumList.Next = nil
	sumList.Val = 0
	if num == 0 {
		return sumList
	}
	tmpNum := num
	for tmpNum != 0 {
		//取个位数
		val := tmpNum % 10
		//去掉个位数
		tmpNum = tmpNum / 10
		//申请链表头(链表头不存储数据)
		newList := new(ListNode)
		newList.Val = val
		newList.Next = nil
		//遍历链表头,找到尾结点
		tmpList := sumList
		for tmpList.Next != nil {
			tmpList = tmpList.Next
		}
		//将新节点尾部插入
		tmpList.Next = newList
	}
	return sumList.Next
}

func main() {
	/**
	*	构造链表 3 ---> 4->3 ---> 2->4->3	342
	 */
	list1 := new(ListNode)
	list1.Val = 3
	list1.Next = nil

	list1 = insert(list1, 4)
	list1 = insert(list1, 2)
	// list := list1
	// for list != nil {
	// 	fmt.Printf("list1.Val: %v\n", list.Val)
	// 	list = list.Next
	// }
	/**
	*	构造链表 4 ---> 6->4 ---> 5->6->4	465
	 */
	list2 := new(ListNode)
	list2.Val = 4
	list2.Next = nil
	list2 = insert(list2, 6)
	list2 = insert(list2, 5)
	//两数相加求和 342+465=807 7->0->8
	listNew := addTwoNumbers(list1, list2)
	print("====================\n")
	list := listNew
	for list != nil {
		fmt.Printf("list1.Val: %v\n", list.Val)
		list = list.Next
	}
}

想法还行,提交失败
未考虑大数相加,例如
输入:
[1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1]
[5,6,4]
输出:
[2,8,0,4,6,2,5,0,3,0,7,2,4,4,9,6,7,0,5]
预期结果:
[6,6,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1]

改进,直接对链表的数字相加进位,返回最长的链表

package main

import "fmt"

type ListNode struct {
	Val  int
	Next *ListNode
}

/**
*	头插入数据 A --->  B->A ---> C->B->A
 */
func insert(head *ListNode, val int) *ListNode {
	newList := new(ListNode)
	newList.Val = val
	newList.Next = head
	return newList
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
	/**
	*	1.判断那个链表更长
	 */
	size1 := 0
	tmp := l1
	for tmp != nil {
		size1++
		tmp = tmp.Next
	}
	fmt.Printf("size1: %v\n", size1)
	size2 := 0
	tmp = l2
	for tmp != nil {
		size2++
		tmp = tmp.Next
	}
	fmt.Printf("size2: %v\n", size2)

	longList := l1
	shotList := l2
	if size1 < size2 {
		longList = l2
		shotList = l1
	}
	//遍历长结点
	tmpLongList := longList
	//进位标志
	carryFlag := 0
	index := 0
	for tmpLongList != nil {
		index++
		num1 := tmpLongList.Val
		num2 := 0
		//遍历短节点
		index2 := 0
		tmpShotList := shotList
		for tmpShotList != nil {
			index2++
			if index2 == index {
				num2 = tmpShotList.Val
				break
			}
			tmpShotList = tmpShotList.Next
		}
		fmt.Printf("num1:%d,num2:%d\n", num1, num2)
		sum := num1 + num2 + carryFlag
		if carryFlag == 1 {
			carryFlag = 0
		}
		if sum >= 10 {
			carryFlag = 1
			sum = sum % 10
		}
		tmpLongList.Val = sum
		// fmt.Printf("sum: %v\n", sum)
		tmpLongList = tmpLongList.Next
	}
	if carryFlag == 1 {
		tmpList := longList
		//找到最好一个节点
		for tmpList.Next != nil {
			tmpList = tmpList.Next
		}
		newListNode := new(ListNode)
		newListNode.Next = nil
		newListNode.Val = 1
		tmpList.Next = newListNode
	}
	fmt.Printf("carryFlag: %v\n", carryFlag)
	return longList
}

func main() {
	/**
	*	构造链表 3 ---> 4->3 ---> 2->4->3	342
	 */
	list1 := new(ListNode)
	list1.Val = 3
	list1.Next = nil

	list1 = insert(list1, 4)
	list1 = insert(list1, 2)
	// list := list1
	// for list != nil {
	// 	fmt.Printf("list1.Val: %v\n", list.Val)
	// 	list = list.Next
	// }
	/**
	*	构造链表 4 ---> 6->4 ---> 5->6->4	465
	 */
	list2 := new(ListNode)
	list2.Val = 4
	list2.Next = nil
	list2 = insert(list2, 6)
	list2 = insert(list2, 5)
	//两数相加求和 342+465=807 7->0->8
	listNew := addTwoNumbers(list1, list2)
	print("====================\n")
	list := listNew
	for list != nil {
		fmt.Printf("list1.Val: %v\n", list.Val)
		list = list.Next
	}
}

首先找到最长的链表和最短的链表,一样长就不用管了
在长链表遍历的循环内遍历短链表,对数字逐个相加得到新的长链表

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