【数据结构系列】链表合并问题——两个链表生成相加链表
题目描述
假设链表中每一个节点的值都在 0 - 9 之间，那么链表整体就可以代表一个整数。
给定两个这种链表，请生成代表两个整数相加值的结果链表。
例如：链表 1 为 9->3->7，链表 2 为 6->3，最后生成新的结果链表为 1->0->0->0。
示例1
输入
[9,3,7],[6,3]
返回值
{1,0,0,0}
解题思路
我的笨办法：
- 利用三个栈模拟两个链表相加，第一个栈存第一个链表的头到尾节点，第二个链表存第二个链表的头到尾节点，由于节点的进位指向头节点，等到两个栈存完节点后取出相加，保留进位，
- 可以把进位随着栈的取出到达前一个节点，把前两个栈取出的值和进位值相加得到的值重新创一个新的ListNode节点存入第三个栈，
- 当前两个栈都取空的时候，第三个栈的节点循环取出重构链表即可得到结果
public class 两个链表生成相加链表 {
public static class ListNode {
int val;
ListNode next = null;
public ListNode(int val) {
this.val = val;
}
}
public ListNode addInList_1(ListNode head1,ListNode head2){
Stack<ListNode> stack1 = new Stack<>(); //存第一个链表的节点
Stack<ListNode> stack2 = new Stack<>(); //存第二个链表的节点
Stack<ListNode> stack3 = new Stack<>(); //存相加后链表的节点
ListNode tail = new ListNode(0);
ListNode result = tail;
if (head1 == null || head2 == null){
if (head1 == null)
return head2;
else return head1;
}
while (head1 != null){
stack1.add(head1);
head1 = head1.next;
}
while (head2 != null){
stack2.add(head2);
head2 = head2.next;
}
int val1=0; //存第一个栈取出的值
int val2=0; //存第二个栈取出的值
int sum =0; //存val1,val2，next的和
int next=0; //定义进位值并初始化
while (!stack1.empty()||!stack2.empty()){
if (!stack1.empty()&&!stack2.empty()){
val1=stack1.pop().val;
val2=stack2.pop().val;
sum=val1+val2+next;
int val=sum%10;
stack3.push(new ListNode(val));
next=sum/10;
}
if (!stack1.empty()&&stack2.empty()){
val1=stack1.pop().val;
sum=val1+next;
int val=sum%10;
stack3.push(new ListNode(val));
next=sum/10;
}
if (stack1.empty()&&!stack2.empty()){
val2=stack2.pop().val;
sum=val2+next;
int val=sum%10;
stack3.push(new ListNode(val));
next=sum/10;
}
}
while (!stack3.empty()){
tail.next= new ListNode(stack3.pop().val);
tail=tail.next;
}
return result.next;
}
public ListNode addInList(ListNode head1,ListNode head2) {
ListNode p1 = Ref(head1);
ListNode p2 = Ref(head2);
ListNode head = new ListNode(0);
int flag = 0;
while(p1!=null&&p2!=null){
head.val = (p1.val + p2.val + flag)%10;
flag = (p1.val + p2.val + flag)/10;
ListNode temp = new ListNode(0);
temp.next = head;
head = temp;
p1 = p1.next;
p2=p2.next;
}
if(p1==null) {
while(p2!=null){
int num = p2.val+flag;
head.val = num%10;
flag = num/10;
ListNode temp = new ListNode(0);
temp.next = head;
head = temp;
p2=p2.next;
}
}
if(p2==null) {
while(p1!=null){
int num = p1.val+flag;
head.val = num%10;
flag = num/10;
ListNode temp = new ListNode(0);
temp.next = head;
head = temp;
p1=p1.next;
}
}
if(flag==0) return head.next;
head.val = flag;
return head;
}
private ListNode Ref(ListNode head) {
if (head == null)
return head;
ListNode p1 = null;
ListNode p2 = null;
ListNode p3 = null;
p2 = head;
p3 = p2.next;
while (p2 != null){
p2.next = p1;
p1 = p2;
p2 = p3;
if (p2 != null)p3 = p3.next;
}
return p1;
}
}
优秀的解法，时间复杂度和空间复杂度大幅度提升了
import java.util.*;
/*
* public class ListNode {
* int val;
* ListNode next = null;
* }
*/
public class Solution {
/**
*
* @param head1 ListNode类
* @param head2 ListNode类
* @return ListNode类
*/
public ListNode addInList (ListNode head1, ListNode head2) {
ListNode p1 = Ref(head1);
ListNode p2 = Ref(head2);
ListNode head = new ListNode(0);
int flag = 0;
while(p1!=null&&p2!=null){
head.val = (p1.val + p2.val + flag)%10;
flag = (p1.val + p2.val + flag)/10;
ListNode temp = new ListNode(0);
temp.next = head;
head = temp;
p1 = p1.next;
p2=p2.next;
}
if(p1==null) {
while(p2!=null){
int num = p2.val+flag;
head.val = num%10;
flag = num/10;
ListNode temp = new ListNode(0);
temp.next = head;
head = temp;
p2=p2.next;
}
}
if(p2==null) {
while(p1!=null){
int num = p1.val+flag;
head.val = num%10;
flag = num/10;
ListNode temp = new ListNode(0);
temp.next = head;
head = temp;
p1=p1.next;
}
}
if(flag==0) return head.next;
head.val = flag;
return head;
}
private static ListNode Ref (ListNode head){
if(head==null)return head;
ListNode p1 = null;
ListNode p2 = null;
ListNode p3 = null;
p2 = head;
p3 = p2.next;
while(p2!=null){
p2.next = p1;
p1 = p2;
p2 = p3;
if(p3!=null)
p3 = p3.next;
}
return p1;
}
}
