标签:des style color java os io for ar
题目:
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { int carry = 0; ListNode res = new ListNode(-1); ListNode head = res; while (l1 != null || l2 != null) { int num = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry; if(l1!=null){ l1 = l1.next; } if(l2!=null){ l2 = l2.next; } res.next = new ListNode(num % 10); res = res.next; carry = num / 10; } if(carry==1){ res.next = new ListNode(1); } return head.next; } }
Input: (3->4->2)+(4->6->5)
Output: (8->0->7)
解析:拓展的这种解法看起来和原题差不多,实际上会更复杂更多,主要要注意两个问题。首先,因为两个链表的长度可能不一样,所以刚开始可以比较两个链表的长度,用0填充较短的链表;原题是求出新的节点后放在旧的节点后面,拓展题目需要添加节点到结果链表的头部,所以用递归调用解决。递归需要传递两个参数:结果链表节点和进位,可以创建内部类包含这两个参数,Java 实现代码如下所示:
/** * Input: (3->4->2)+(4->6->5) Output: (8->0->7) */ public class AddTwoNumbersExp { public static void main(String[] args) { ListNode node11 = new ListNode(3); ListNode node12 = new ListNode(4); ListNode node13 = new ListNode(2); ListNode node21 = new ListNode(4); ListNode node22 = new ListNode(6); ListNode node23 = new ListNode(5); node11.next = node12; node12.next = node13; node21.next = node22; node22.next = node23; ListNode res = new AddTwoNumbersExp().addTwoNumbers(node11, node21); while (res != null) { System.out.print(res.val+" "); res = res.next; } } public class Node { public ListNode sum = null; int carry; } public ListNode addTwoNumbers(ListNode l1, ListNode l2) { // 查询两个链表长度,将短的链表用0填充 int len1 = getLength(l1); int len2 = getLength(l2); l1 = len1 < len2 ? paddingZero(l1, len2 - len1) : l1; l2 = len2 < len1 ? paddingZero(l2, len1 - len2) : l2; // 对两个链表求和 Node sum = addTwoList(l1, l2); // 判断是否有进位 if (sum.carry == 0) { return sum.sum; } ListNode node = new ListNode(1); node.next = sum.sum; return node; } public int getLength(ListNode node) { int length = 0; while (node != null) { length++; node = node.next; } return length; } public ListNode paddingZero(ListNode l1, int zeroN) { ListNode res = new ListNode(-1); res.next = l1; for (int i = 0; i < zeroN; i++) { ListNode node = new ListNode(0); node.next = res.next; res.next = node; } return res.next; } public Node addTwoList(ListNode l1, ListNode l2) { if (l1 == null && l2 == null) { Node res = new Node(); return res; } Node after = addTwoList(l1.next, l2.next); int val = after.carry + l1.val + l2.val; ListNode cur = new ListNode(val % 10); cur.next = after.sum; Node res = new Node(); res.carry = val / 10; res.sum = cur; return res; } }
【leetcode】Add Two Numbers 解析以及拓展,布布扣,bubuko.com
【leetcode】Add Two Numbers 解析以及拓展
标签:des style color java os io for ar
原文地址:http://blog.csdn.net/u013378502/article/details/38467895