LintCode | 167. 链表求和

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
public ListNode addLists(ListNode l1, ListNode l2) {
//先排除两个链表均为null的情况
if(l1 == null && l2 == null) return null;

//carry表示每次相加的进位
int carry = 0;

//result用于保存头节点，temp用于循环，pre用于循环开始后保存temp的上一个节点
ListNode result = new ListNode(0);
ListNode temp = result;
ListNode pre = temp;

//当两个链表都循环到最后一个节点且没有进位项时结束循环
while(l1 != null || l2 != null || carry != 0) {
//先计算得同一位置的节点与进位的总和
if(l1 != null) {
temp.val = temp.val + l1.val;
l1 = l1.next;
}
if(l2 != null) {
temp.val = temp.val + l2.val;
l2 = l2.next;
}
temp.val = temp.val + carry;

//将和的进位项另外保存，保留个位
carry = temp.val / 10;
temp.val = temp.val - carry * 10;
temp.next = new ListNode(0);
pre = temp;
temp = temp.next;
}
pre.next = null;
return result;
}
}