add-two-numbers-ii
注意:有一种好的方法,是将链表倒转,然后依次相加。
但是,按照题目要求,用了不改变原链表的方法。
就是将两个链表增加到相同长度,然后递归相加,子函数返回后处理进位。
https://leetcode.com/problems/add-two-numbers-ii/
package com.company; import java.util.*; class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
class Solution { ListNode addTwo(ListNode l1, ListNode l2) {
//System.out.printf("add two %d, %d \n", l1.val, l2.val);
ListNode ret = new ListNode(l1.val + l2.val);
if (l1.next != null && l2.next != null) {
ret.next = addTwo(l1.next, l2.next);
ret.val += ret.next.val / 10;
ret.next.val = ret.next.val % 10;
}
else {
ret.next = null;
}
return ret;
} public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int l1len = 0;
ListNode ln = l1;
while (ln != null) {
l1len++;
ln = ln.next;
}
int l2len = 0;
ln = l2;
while (ln != null) {
l2len++;
ln = ln.next;
}
if (l1len < l2len) {
ln = l1;
l1 = l2;
l2 = ln;
l1len = l1len ^ l2len;
l2len = l1len ^ l2len;
l1len = l1len ^ l2len;
} for (int i=0; i<l1len-l2len; i++) {
ln = new ListNode(0);
ln.next = l2;
l2 = ln;
} ln = addTwo(l1, l2);
if (ln.val >= 10) {
ListNode newHead = new ListNode(ln.val / 10);
ln.val = ln.val % 10;
newHead.next = ln;
ln = newHead;
}
return ln;
}
} public class Main { public static void main(String[] args) {
System.out.println("Hello!");
Solution solution = new Solution(); ListNode l1 = new ListNode(7);
ListNode l11 = new ListNode(2);
ListNode l12 = new ListNode(4);
ListNode l13 = new ListNode(3);
l1.next = l11;
l11.next = l12;
l12.next = l13;
ListNode l2 = new ListNode(5);
ListNode l21 = new ListNode(6);
ListNode l22 = new ListNode(4);
l2.next = l21;
l21.next = l22;
ListNode ret = solution.addTwoNumbers(l1, l2);
System.out.printf("Get ret: \n");
while (ret != null) {
System.out.printf("%d", ret.val);
ret = ret.next;
}
System.out.println(); /*Iterator<List<Integer>> iterator = ret.iterator();
while (iterator.hasNext()) {
Iterator iter = iterator.next().iterator();
while (iter.hasNext()) {
System.out.printf("%d,", iter.next());
}
System.out.println();
}*/ System.out.println(); }
}
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