LintCode:链表求和
2017-07-25 20:25
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描述:
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
样例:
给出两个链表
代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
ListNode *addLists(ListNode *l1, ListNode *l2) {
// write your code here
int add_Bit = 0;
ListNode *m_head,*p;
m_head = (ListNode*)malloc(sizeof(ListNode));
p = m_head;
p -> next = NULL;
p -> val = 0;
while(l1||l2){
if(l1 && l2){
p -> val = (l1 -> val + l2 -> val + add_Bit) % 10;
add_Bit = (l1 -> val + l2 -> val + add_Bit) / 10;
l1 = l1 -> next;
l2 = l2 -> next;
}
else if(l1 && !l2){
p -> val = (l1 -> val + add_Bit) % 10;
add_Bit = (l1 -> val + add_Bit) / 10;
l1 = l1 -> next;
}
else if(!l1 && l2){
90bc
p -> val = (l2 -> val + add_Bit) % 10;
add_Bit = (l2 -> val + add_Bit) / 10;
l2 = l2 -> next;
}
p -> next = (ListNode*)malloc(sizeof(ListNode));
if(l1||l2){
p = p -> next;
}
}
if(add_Bit){
p = p -> next;
p -> val = 1;
}
p -> next = NULL;
return m_head;
}
};
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。
样例:
给出两个链表
3->1->5->null和
5->9->2->null,返回
8->0->8->null
代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
ListNode *addLists(ListNode *l1, ListNode *l2) {
// write your code here
int add_Bit = 0;
ListNode *m_head,*p;
m_head = (ListNode*)malloc(sizeof(ListNode));
p = m_head;
p -> next = NULL;
p -> val = 0;
while(l1||l2){
if(l1 && l2){
p -> val = (l1 -> val + l2 -> val + add_Bit) % 10;
add_Bit = (l1 -> val + l2 -> val + add_Bit) / 10;
l1 = l1 -> next;
l2 = l2 -> next;
}
else if(l1 && !l2){
p -> val = (l1 -> val + add_Bit) % 10;
add_Bit = (l1 -> val + add_Bit) / 10;
l1 = l1 -> next;
}
else if(!l1 && l2){
90bc
p -> val = (l2 -> val + add_Bit) % 10;
add_Bit = (l2 -> val + add_Bit) / 10;
l2 = l2 -> next;
}
p -> next = (ListNode*)malloc(sizeof(ListNode));
if(l1||l2){
p = p -> next;
}
}
if(add_Bit){
p = p -> next;
p -> val = 1;
}
p -> next = NULL;
return m_head;
}
};
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