链表求和
2017-05-23 14:59
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样例
给出 6->1->7 + 2->9->5。即,617 + 295。
返回 9->1->2。即,912 。
/**
* 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 addLists2(ListNode l1, ListNode l2) {
l1 = ReverseList(l1);
l2 = ReverseList(l2);
int tmp = 0;
ListNode node = new ListNode(0);
ListNode p = node;
while (tmp != 0 || l1 != null || l2 != null) {
if (l1 != null) {
tmp += l1.val;
l1 = l1.next;
}
if (l2 != null) {
tmp += l2.val;
l2 = l2.next;
}
p.next = new ListNode(tmp % 10);
p = p.next;
tmp = tmp / 10;
}
return ReverseList(node.next);
}
}
给出 6->1->7 + 2->9->5。即,617 + 295。
返回 9->1->2。即,912 。
/**
* 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 addLists2(ListNode l1, ListNode l2) {
l1 = ReverseList(l1);
l2 = ReverseList(l2);
int tmp = 0;
ListNode node = new ListNode(0);
ListNode p = node;
while (tmp != 0 || l1 != null || l2 != null) {
if (l1 != null) {
tmp += l1.val;
l1 = l1.next;
}
if (l2 != null) {
tmp += l2.val;
l2 = l2.next;
}
p.next = new ListNode(tmp % 10);
p = p.next;
tmp = tmp / 10;
}
return ReverseList(node.next);
}
public static ListNode ReverseList(ListNode head) { if(head == null) return null; ListNode p = head; ListNode q = p.next; p.next = null; while(q != null) { head = q.next; q.next = p; p = q; q = head; } head = p; return head; }
}
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