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Solution_2.java
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81 lines (80 loc) · 2.61 KB
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/*
2. 两数相加
给出两个非空的链表用来表示两个非负的整数。其中,它们各自的位数是按照逆序的方式存储的,并且它们的每个节点只能存储一位字。
如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外,这两个数都不会以 0 开头。
示例:
输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出:7 -> 0 -> 8
原因:342 + 465 = 807
*/
class Solution {
public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode l;
int addition;
boolean flag;
addition = l1.val + l2.val;
if (addition >= 10) {
addition -= 10;
flag = true;
} else flag = false;
l = new ListNode(addition);
l1 = l1.next;
l2 = l2.next;
ListNode result = l;
while ( l1 != null && l2 != null) {
if (flag) l.next = new ListNode(1);
else l.next = new ListNode(0);
addition = l1.val + l2.val;
if (addition >= 10) {
addition -= 10;
flag = true;
} else flag = false;
l.next.val += addition;
if ( l.next.val >= 10 ) {
flag = true;
l.next.val -= 10;
}
l1 = l1.next;
l2 = l2.next;
l = l.next;
}
if (l1 == null && l2 != null) {
while (l2!=null) {
if (flag == true) {
l.next = new ListNode(1);
l.next.val += l2.val;
if ( l.next.val >= 10 ) {
l.next.val -= 10;
flag =true;
} else flag = false;
} else l.next = new ListNode(l2.val);
l = l.next;
l2 = l2.next;
}
}
if (l1 != null && l2 == null) {
while (l1 != null) {
if (flag == true) {
l.next = new ListNode(1);
l.next.val += l1.val;
if ( l.next.val >= 10 ) {
l.next.val -= 10;
flag =true;
} else flag = false;
} else l.next = new ListNode(l1.val);
l = l.next;
l1 = l1.next;
}
}
if (l1 == null && l2 == null) {
if (flag == true)
l.next = new ListNode(1);
}
if ( l.val >= 10 ) {
l.next = new ListNode(1);
l.val -= 10;
}
return result;
}
}