2. Add Two Numbers （两数求和）

布满荆棘的人生 2022-06-12 02:15 143阅读 0赞

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

**题目大意：两个非负数分别逆序存放在两个单链表中，将两个单链表代表的数求和，并将和以单链表的形式返回。**

**解题思路：双指针，先遍历完较短的链表，将两个链表对应节点值求和之后的值更新到较长的链表中，较短的链表遍历完后再接着遍历较长链表剩下的部分，此时仅以较长链表中节点的值作为求和加数，但是要注意处理有进位的情况。**

**源代码：（57 ms，beats 47.31%）**

/**
     * 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) {
            int lenth1 = 0, lenth2 = 0;
    		ListNode p = l1, q = l2, result;
    		int curRes;
    		// 进位
    		int carry = 0;
    		while (p != null) {
    			lenth1++;
    			p = p.next;
    		}
    		while (q != null) {
    			lenth2++;
    			q = q.next;
    		}
    		if (lenth1 < lenth2) {
    			result = l2;
    			p = l1;
    			q = l2;
    		} else {
    			result = l1;
    			p = l2;
    			q = l1;
    		}
    		ListNode preq = null;
    		while (p != null) {
    			curRes = p.val + q.val + carry;
    			if (curRes >= 10) {
    				carry = 1;
    				curRes = curRes % 10;
    			} else {
    				carry = 0;
    			}
    			q.val = curRes;
    			preq = q;
    			p = p.next;
    			q = q.next;
    		}
    		while (q != null) {
    			curRes = q.val + carry;
    			if (curRes >= 10) {
    				carry = 1;
    				curRes = curRes % 10;
    			} else {
    				carry = 0;
    			}
    			q.val = curRes;
    			preq = q;
    			q = q.next;
    		}
    		if (carry == 1) {
    			preq.next = new ListNode(1);
    		}
    		return result;
        }
    }