两个链表生成相加链表
题目：
思路：
本题难点：第一，不知道链条有多大，最合起来的数可能会远大于long与int能存的极限。
思路一，将链表反转，链条反转的链表从个位数开始相加，然后取余数，不断叠成新的链表。
思路二，将链表的数据取出组合成字符串，然后利用大数相加的方法，取得相加的和，然后根据新的和的字符串生成新的链表。（但实际上运算量有点过大，没有过测试）
代码示例：
public class Solution4 {
public static void main(String[] args) {
ListNode head = new ListNode(6);
ListNode node2 = new ListNode(3);
ListNode node3 = new ListNode(9);
ListNode node4 = new ListNode(3);
ListNode node5 = new ListNode(7);
head.next = node2;
node3.next = node4;
node4.next = node5;
ListNode result = addInList(head, node3);
System.out.println(result);
OutPutLinkedList(result);
}
/**
* 方案2，采用合成字符串的形式
* 但是这种方式运算效率太低了
*
* @param head1 ListNode类
* @param head2 ListNode类
* @return ListNode类
*/
public static ListNode addInList(ListNode head1, ListNode head2) {
// write code here
if (head1 == null)
return head2;
if (head2 == null)
return head1;
ListNode temp1 = head1, temp2 = head2;
String str1 = "", str2 = "";
while (temp1 != null) {
str1 = str1 + temp1.val;
temp1 = temp1.next;
}
while (temp2 != null) {
str2 = str2 + temp2.val;
temp2 = temp2.next;
}
String temp = solve(str1, str2);
//构建头结点
ListNode head = new ListNode(temp.charAt(0) - '0');
ListNode result = head;
for (int i = 1; i < temp.length(); i++) {
ListNode tempNode = new ListNode((int) temp.charAt(i) - '0');
head.next = tempNode;
head = tempNode;
}
return result;
}
public static String solve(String s, String t) {
StringBuilder ans = new StringBuilder();
int tmp = 0;
int ls = s.length() - 1, lt = t.length() - 1;
while (ls >= 0 || lt >= 0 || tmp == 1) {
int l = ls >= 0 ? (s.charAt(ls--) - '0') : 0;
int r = lt >= 0 ? (t.charAt(lt--) - '0') : 0;
int plus = l + r + tmp;
tmp = plus / 10;
char a = (char) (plus % 10 + '0');
ans.append(a);
}
return ans.reverse().toString();
}
/**
* 方案1，采用反转链表的形式
*
* @param head1 ListNode类
* @param head2 ListNode类
* @return ListNode类
*/
public static ListNode addInList1(ListNode head1, ListNode head2) {
// write code here
if (head1 == null)
return head2;
if (head2 == null)
return head1;
int num, num1, num2;
//颠倒两个链条
ListNode temp1 = reverseIteratively(head1);
ListNode temp2 = reverseIteratively(head2);
//构建头结点
num = temp1.val + temp2.val;
temp1 = temp1.next;
temp2 = temp2.next;
ListNode head = new ListNode(num % 10);
num /= 10;
//利用遍历产生链条
while (temp1 != null || temp2 != null || num != 0) {
num1 = 0;
num2 = 0;
if (temp1 != null) {
num1 = temp1.val;
temp1 = temp1.next;
}
if (temp2 != null) {
num2 = temp2.val;
temp2 = temp2.next;
}
num = num + num1 + num2;
ListNode temp = new ListNode(num % 10);
temp.next = head;
head = temp;
num /= 10;
}
return head;
}
public static ListNode reverseIteratively(ListNode head) {
ListNode pre = null;
ListNode next = null;
while (head != null) {
next = head.next;
head.next = pre;
pre = head;
head = next;
}
return pre;
}
/**
* 打印链表，用于输出查看
*
* @param head
*/
public static void OutPutLinkedList(ListNode head) {
ListNode temp = head;
while (null != temp) {
System.out.print(temp.val);
if (null != temp.next) {
System.out.print(" -> ");
}
temp = temp.next;
}
System.out.println();
}
}
class ListNode {
int val;
ListNode next;
ListNode() {
}
ListNode(int x) {
val = x;
next = null;
}
}
