167. Add Two Numbers【easy】
You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse
order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.
Given 7->1->6 + 5->9->2
. That is, 617 + 295
.
Return 2->1->9
. That is 912
.
Given 3->1->5
and 5->9->2
, return 8->0->8
.
题意
你有两个用链表代表的整数,其中每个节点包含一个数字。数字存储按照在原来整数中相反
的顺序,使得第一个数字位于链表的开头。写出一个函数将两个整数相加,用链表形式返回和。
解法一:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode addLists(ListNode l1, ListNode l2) {
if(l1 == null && l2 == null) {
return null;
} ListNode head = new ListNode(0);
ListNode point = head;
int carry = 0;
while(l1 != null && l2!=null){
int sum = carry + l1.val + l2.val;
point.next = new ListNode(sum % 10);
carry = sum / 10;
l1 = l1.next;
l2 = l2.next;
point = point.next;
} while(l1 != null) {
int sum = carry + l1.val;
point.next = new ListNode(sum % 10);
carry = sum /10;
l1 = l1.next;
point = point.next;
} while(l2 != null) {
int sum = carry + l2.val;
point.next = new ListNode(sum % 10);
carry = sum /10;
l2 = l2.next;
point = point.next;
} if (carry != 0) {
point.next = new ListNode(carry);
}
return head.next;
}
}
中规中矩的解法
解法二:
public class Solution {
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
public ListNode addLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode tail = dummy; int carry = 0;
for (ListNode i = l1, j = l2; i != null || j != null; ) {
int sum = carry;
sum += (i != null) ? i.val : 0;
sum += (j != null) ? j.val : 0; tail.next = new ListNode(sum % 10);
tail = tail.next; carry = sum / 10;
i = (i == null) ? i : i.next;
j = (j == null) ? j : j.next;
} if (carry != 0) {
tail.next = new ListNode(carry);
}
return dummy.next;
}
}
比较简明的写法,且使用了dummy节点,参考@NineChapter 的代码
解法三:
public class Solution {
/*
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
public ListNode addLists(ListNode l1, ListNode l2) {
ListNode root = new ListNode(-1);
ListNode result = root;
int carry = 0; while( l1 != null || l2 != null || carry == 1){
int value = 0;
if(l1 != null){
value += l1.val;
l1 = l1.next;
}
if( l2 != null){
value += l2.val;
l2 = l2.next;
} value += carry;
root.next = new ListNode(value % 10);
carry = value / 10; root = root.next; } return result.next;
}
}
解法四:
class Solution {
public:
ListNode* addLists(ListNode* l1, ListNode* l2) {
ListNode *head = new ListNode();
ListNode *ptr = head;
int carry = ;
while (true) {
if (l1 != NULL) {
carry += l1->val;
l1 = l1->next;
}
if (l2 != NULL) {
carry += l2->val;
l2 = l2->next;
}
ptr->val = carry % ;
carry /= ;
// 当两个表非空或者仍有进位时需要继续运算,否则退出循环
if (l1 != NULL || l2 != NULL || carry != ) {
ptr = (ptr->next = new ListNode());
} else {
break;
}
}
return head;
}
};
非常简明的代码,参考@NineChapter 的代码
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