算法练习2.Add Two Numbers 反向小数链表表示相加
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2. Add Two Numbers
https://leetcode.com/problems/add-two-numbers/
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
/**
*Definition for singly-linked list.
*struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
反向小数链表表示相加(先是小数点最后一位,之后链表反向表示,最后一个是个位)class Solution {
public:
ListNode*addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode*add = new ListNode(1);
ListNode*temp = add;
ints=0;
while(l1&&l2)
{
if(s + l1->val + l2->val >= 10)
{
add->val= s + l1->val + l2->val-10;
s= 1;
}
else{
add->val= s + l1->val + l2->val;
s= 0;
}
l1= l1->next;
l2= l2->next;
if(l1&&l2)
{
ListNode*n = new ListNode(1);
add->next= n;
add= add->next;
}
}
while(l1)
{
ListNode *n = new ListNode(1);
add->next = n;
add = add->next;
if (s + l1->val >= 10)
{
add->val= s + l1->val-10;
s= 1;
}
else{
add->val= s + l1->val;
s= 0;
}
l1= l1->next;
}
while(l2)
{
ListNode *n = new ListNode(1);
add->next = n;
add = add->next;
if (s + l2->val >= 10)
{
add->val= s + l2->val-10;
s= 1;
}
else{
add->val= s + l2->val;
s= 0;
}
l2= l2->next;
}
if(s==1)
{
ListNode *n = new ListNode(1);
add->next = n;
}
returntemp;
}
};
以下内容转自LeetCode官网:
public ListNodeaddTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(0);
ListNode p = l1, q= l2, curr=dummyHead;
int carry=0;
while (p != null || q != null){
int x= (p!= null) ? p.val:0;
int y= (q!= null) ? q.val:0;
int sum= carry+ x + y;
carry = sum / 10;
curr.next= new ListNode(sum% 10);
curr = curr.next;
if (p != null) p = p.next;
if (q != null) q = q.next;
}
if (carry > 0){
curr.next= new ListNode(carry);
}
return dummyHead.next;
}
参考官网解答后修改版:
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1,ListNode* l2) {
ListNode *add = new ListNode(0);
ListNode *temp = add;
int s = 0, x, y, sum;
while (l1 != NULL || l2 != NULL)
{
x = (l1 != NULL) ? l1->val : 0;
y = (l2 != NULL) ? l2->val : 0;
sum = s + x + y;
s = sum / 10;
add->next = new ListNode(sum % 10);
add = add->next;
if (l1!= NULL) l1 = l1->next;
if (l2!= NULL) l2 = l2->next;
}
if (s>0)
{
add->next = new ListNode(s);
}
return temp->next;
}
};
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