題目描述
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
我的解答
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry = 0;
ListNode newHead = new ListNode(0);
ListNode p1 = l1,p2=l2,p3=newHead;
while (p1 != null || p2 != null){
if(p1 != null){
carry += p1.val;
p1 = p1.next;
}
if(p2 != null){
carry += p2.val;
p2 = p2.next;
}
p3.next = new ListNode(carry%10);
carry /= 10;
p3 = p3.next;
}
if(carry == 1){
p3.next = new ListNode(1);
}
return newHead.next;
}
public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
分析
這道題比較容易渣蜗，為簡(jiǎn)單鏈表題屠尊，主要用carry來(lái)維護(hù)所加的結(jié)果，需要注意的情況是當(dāng)一個(gè)鏈表已經(jīng)結(jié)束而另一個(gè)還未結(jié)束的情況耕拷，還有全部遍歷完之后看下carry是否為1讼昆，如果為1要在后面多加一位。
