LeetCode2 AddTowNumbers

标签：leetcode   算法   链表   

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

译：有两个存储非负整数的链表，数字的每一位以逆序的方式存储在链表中的，计算两个数的值，返回对应的链表。

方法：和链表的归并排序算法类似，注意处理进位。

public static class ListNode {
		int val;
		ListNode next; 
		ListNode(int x) {
			val = x;
		}
	}

	public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		int a = 0;//表示进位
		ListNode head = null;
		ListNode tail = head;
		ListNode list1 = l1;
		ListNode list2 = l2;
		while (list1 != null && list2 != null) {
			ListNode t = new ListNode((list1.val + list2.val + a) % 10);
			if (head == null) {
				head = t;
				tail = head;
			} else {
				tail.next = t;
				tail = t;
			}
			a = (list1.val + list2.val + a) / 10;
			list1 = list1.next;
			list2 = list2.next;
		}

		while (list1 != null) {//如果List1表示的数更大,处理List1的高位
			ListNode t = new ListNode((list1.val + a) % 10);
			tail.next = t;
			tail = t;
			a = (list1.val + a) / 10;
			list1 = list1.next;
		}

		while (list2 != null) {//如果List2表示的数更大，处理List2的高位
			ListNode t = new ListNode((list2.val + a) % 10);
			tail.next = t;
			tail = t;
			a = (list2.val + a) / 10;
			list2 = list2.next;
		}

		if (a > 0) {//处理进位
			ListNode t = new ListNode(a);
			tail.next = t;
			tail = t;
		}
		return head;
	}

LeetCode2 AddTowNumbers

标签：leetcode   算法   链表   

原文地址：http://blog.csdn.net/zhang1314wen2008/article/details/46387897