leetcode2. Add Two Numbers
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
解法一
每一位相加,大于10向后加1,依次添加节点
/**
* @Author RenXintao
* @Date 2/10/17
*/
class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
}
}
public class Solution2 {
public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if(l1 == null && l2 == null) {
return null;
}
ListNode result = new ListNode(0);
ListNode head = null;
int sum, flag = 0;
while(l1 != null || l2 != null || flag != 0) {
// l2节点为空
if(l1 != null && l2 == null) {
sum = l1.val + flag;
// l1节点为空
} else if (l1 == null && l2 != null) {
sum = l2.val + flag;
// l1和l2节点都为空,前一项的和>9
} else if (l1 == null && l2 == null && flag != 0) {
sum = flag;
// 都不为空
} else {
sum = l1.val + l2.val + flag;
}
// 两项和>9,flag为1;否则为0;
if(sum > 9) {
sum = sum % 10;
flag = 1;
} else {
flag = 0;
}
// 链表中添加节点
if (head != null) {
head.next = new ListNode(sum);
head = head.next;
} else {
head = new ListNode(sum);
result = head;
}
// 遍历l1节点
if(l1 != null) {
l1 = l1.next;
}
// 遍历l2节点
if(l2 != null) {
l2 = l2.next;
}
}
return result;
}
public static void main(String[] args) {
ListNode head1 = new ListNode(2);
ListNode l1 = head1;
head1.next = new ListNode(4);
head1 = head1.next;
head1.next = new ListNode(3);
ListNode head2 = new ListNode(5);
ListNode l2 = head2;
head2.next = new ListNode(6);
head2 = head2.next;
head2.next = new ListNode(7);
ListNode ret = addTwoNumbers(l1, l2);
while (ret != null) {
System.out.println(ret.val);
ret = ret.next;
}
}
}
结果:
7
0
1
1
本文介绍了解决LeetCode题目“两数相加”的一种方法。该方法通过链表逆序存储两个非负整数,并逐位相加形成新的链表,当和大于10时向前一位进1。
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