https://leetcode-cn.com/problems/add-two-numbers/submissions/
我的方法:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode ans = new ListNode(0);
ans.next = null;
ListNode curr = ans; int c = 0;
while(l1.next != null && l2.next != null){
curr.val = (l1.val + l2.val + c)%10;
c = (l1.val + l2.val + c)/10;
curr.next = new ListNode(0);
curr = curr.next;
l1 = l1.next;
l2 = l2.next;
}
//两数等长度时同时处理最后一个数即可
if(l1.next == null && l2.next == null){
curr.val = (l1.val + l2.val + c)%10;
c = (l1.val + l2.val +c)/10;
//看最终是否还有一位进位
if(c == 1){
curr.next = new ListNode(c);
curr.next.next = null;
}else{
curr.next = null;
}
return ans;
}
//一长一短时,留下长的那个,把短的的最后一个val记录后用不着了
ListNode l;
if(l1.next != null && l2.next == null){
l = l1;
curr.val = l2.val;
}else{
l = l2;
curr.val = l1.val;
}
//处理短的的最后一位
curr.val += l.val + c;
c = curr.val / 10;
curr.val %= 10;
curr.next = new ListNode(0);
curr = curr.next;
l = l.next; while(l.next != null){
curr.val = (l.val + c)%10;
c = (l.val + c)/10;
curr.next = new ListNode(0);
curr = curr.next;
l = l.next;
} //处理长的的最后一位
curr.val = (l.val + c)%10;
c = (l.val + c)/10;
if(c == 1){
curr.next = new ListNode(c);
curr.next.next = null;
}else{
curr.next = null;
} return ans;
}
}
按部就班的考虑与处理.注意不要有考察漏掉的情况
时间复杂度O(n),空间复杂度O(n)
//我所有的时空复杂度都是指级别,有必要具体分析的会特别注明
结果:
官方题解方法:
https://leetcode-cn.com/problems/add-two-numbers/solution/liang-shu-xiang-jia-by-leetcode/
也是朴素方法,区别在于他没有拆开来考虑,而是用了或判断,在末尾前进的时候又加了if
代码量比我的少,优雅一些.
(见仁见智,我不喜欢在简单的循环里加判断,徒增复杂度)
