leetcode之Merge Two Sorted Lists
2013-09-05 15:48
375 查看
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (l1 == NULL)
return l2;
if (l2 == NULL)
return l1;
ListNode* head;
ListNode* pos1 = l1;
ListNode* pos2 = l2;
if (l1->val > l2->val) {
head = l2;
pos2 = l2->next;
} else {
head = l1;
pos1 = pos1->next;
}
ListNode* pos = head;
while (pos1!=NULL && pos2!=NULL) {
if (pos1->val <= pos2->val) {
pos->next = pos1;
pos1=pos1->next;
} else {
pos->next = pos2;
pos2=pos2->next;
}
pos = pos->next;
}
if (pos1 == NULL&&pos2==NULL) {
return head;
} else if (pos1 !=NULL) {
pos->next = pos1;
} else {
pos->next = pos2;
}
return head;
}
};
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (l1 == NULL)
return l2;
if (l2 == NULL)
return l1;
ListNode* head;
ListNode* pos1 = l1;
ListNode* pos2 = l2;
if (l1->val > l2->val) {
head = l2;
pos2 = l2->next;
} else {
head = l1;
pos1 = pos1->next;
}
ListNode* pos = head;
while (pos1!=NULL && pos2!=NULL) {
if (pos1->val <= pos2->val) {
pos->next = pos1;
pos1=pos1->next;
} else {
pos->next = pos2;
pos2=pos2->next;
}
pos = pos->next;
}
if (pos1 == NULL&&pos2==NULL) {
return head;
} else if (pos1 !=NULL) {
pos->next = pos1;
} else {
pos->next = pos2;
}
return head;
}
};
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