LEETCODE: Add Two Number
2014-12-04 09:43
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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.
第一感觉,和归并排序中的merge操作有点相似。
归并排序的时候,在循环合并两个有序序列的时候,有三种情况:
1. 两个序列都有值,找出那个小的插入新的序列;
2.只有序列1有值,把它插入新的序列;
2.只有序列2有值,把它插入新的序列。
对于这个例子,情况稍微复杂一点,有四种情况:
1.两个序列都有值,把当前节点上的值相加,如果有进位,还有加上进位,同时,如果和大于10,说有有进位,设置进位标志。将结果加入新的序列。
2.只有序列1有值,把当前节点的值和进位值相加,如果和大于10,说有有进位,设置进位标志。将结果加入新的序列。
3.只有序列2有值,同上。
4.两个序列都已经没有值了,查看是否有进位,如果有进位,需要将进位也插入新的序列。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode *returnHead = NULL;
ListNode *currentNode = NULL;
int increase = 0;
while(l1 != NULL && l2 != NULL)
{
ListNode *newHead = new ListNode((l1->val + l2->val + increase) % 10);
increase = (l1->val + l2->val + increase) / 10;
if(currentNode == NULL)
{
returnHead = newHead;
}
else
{
currentNode->next = newHead;
}
l1 = l1->next;
l2 = l2->next;
currentNode = newHead;
}
while(l1 != NULL)
{
ListNode *newHead = new ListNode((l1->val + increase) % 10);
increase = (l1->val + increase) / 10;
if(currentNode == NULL)
{
returnHead = newHead;
}
else
{
currentNode->next = newHead;
}
l1 = l1->next;
currentNode = newHead;
}
while(l2 != NULL)
{
ListNode *newHead = new ListNode((l2->val + increase) % 10);
increase = (l2->val + increase) / 10;
if(currentNode == NULL)
{
returnHead = newHead;
}
else
{
currentNode->next = newHead;
}
l2 = l2->next;
currentNode = newHead;
}
if(increase > 0)
{
ListNode *newHead = new ListNode(increase);
if(currentNode == NULL)
{
returnHead = newHead;
}
else
{
currentNode->next = newHead;
}
}
return returnHead;
}
};
第一感觉,和归并排序中的merge操作有点相似。
归并排序的时候,在循环合并两个有序序列的时候,有三种情况:
1. 两个序列都有值,找出那个小的插入新的序列;
2.只有序列1有值,把它插入新的序列;
2.只有序列2有值,把它插入新的序列。
对于这个例子,情况稍微复杂一点,有四种情况:
1.两个序列都有值,把当前节点上的值相加,如果有进位,还有加上进位,同时,如果和大于10,说有有进位,设置进位标志。将结果加入新的序列。
2.只有序列1有值,把当前节点的值和进位值相加,如果和大于10,说有有进位,设置进位标志。将结果加入新的序列。
3.只有序列2有值,同上。
4.两个序列都已经没有值了,查看是否有进位,如果有进位,需要将进位也插入新的序列。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode *returnHead = NULL;
ListNode *currentNode = NULL;
int increase = 0;
while(l1 != NULL && l2 != NULL)
{
ListNode *newHead = new ListNode((l1->val + l2->val + increase) % 10);
increase = (l1->val + l2->val + increase) / 10;
if(currentNode == NULL)
{
returnHead = newHead;
}
else
{
currentNode->next = newHead;
}
l1 = l1->next;
l2 = l2->next;
currentNode = newHead;
}
while(l1 != NULL)
{
ListNode *newHead = new ListNode((l1->val + increase) % 10);
increase = (l1->val + increase) / 10;
if(currentNode == NULL)
{
returnHead = newHead;
}
else
{
currentNode->next = newHead;
}
l1 = l1->next;
currentNode = newHead;
}
while(l2 != NULL)
{
ListNode *newHead = new ListNode((l2->val + increase) % 10);
increase = (l2->val + increase) / 10;
if(currentNode == NULL)
{
returnHead = newHead;
}
else
{
currentNode->next = newHead;
}
l2 = l2->next;
currentNode = newHead;
}
if(increase > 0)
{
ListNode *newHead = new ListNode(increase);
if(currentNode == NULL)
{
returnHead = newHead;
}
else
{
currentNode->next = newHead;
}
}
return returnHead;
}
};
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