标签:turn 简单 content link ddt 数字 val 逻辑 des
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
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { boolean flag = false; ListNode res = new ListNode(0); ListNode head = res; while (l1 != null && l2 != null) { int digit = l1.val + l2.val; ListNode d = null; if (digit > 9) { d = flag ? new ListNode(digit % 10 + 1) : new ListNode(digit % 10); flag = true; } else if (flag) { if (digit + 1 > 9) { d = new ListNode((digit + 1) % 10); flag = true; } else { d = new ListNode(digit + 1); flag = false; } } else { d = new ListNode(digit); flag = false; } res.next = d; res = d; l1 = l1.next; l2 = l2.next; } while (l1 != null) { int digit = 0; if (flag) { digit = l1.val + 1; if (digit > 9) { flag = true; digit = digit % 10; } else { flag = false; } } else { digit = l1.val; flag = false; } ListNode d = new ListNode(digit); res.next = d; res = d; l1 = l1.next; } while (l2 != null) { int digit = 0; if (flag) { digit = l2.val + 1; if (digit > 9) { flag = true; digit = digit % 10; } else { flag = false; } } else { digit = l2.val; flag = false; } ListNode d = new ListNode(digit); res.next = d; res = d; l2 = l2.next; } if (flag) { ListNode d = new ListNode(1); res.next = d; } return head.next; } }
我是把相同的位数相加,然后再判断剩下那个比较长的数字。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode res = new ListNode(0); ListNode p = res; int sum = 0; while (l1 != null || l2 != null) { if (l1 != null) { sum += l1.val; l1 = l1.next; } if (l2 != null) { sum += l2.val; l2 = l2.next; } p.next = new ListNode(sum % 10); p = p.next; sum /= 10; } if (sum == 1) { p.next = new ListNode(1); } return res.next; } }
反正一对比就高下立现。。。嘤嘤嘤。。。
标签:turn 简单 content link ddt 数字 val 逻辑 des
原文地址:http://www.cnblogs.com/aprilyang/p/6349198.html