LeetCode #021 MergeTwoSortedLists. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
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LeetCode #21
Q.
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
두개의 정렬된 연결리스트를 병합해서 새로운 리스트를 반환해라. 새로운 리스트는 처음 두개의 리스트 노드들을 잘라낸 것으로 만들어져야 한다.
e.g.
Example:
Input: 1->2->4 , 1->3->4
Output: 1->1->2->3->4->4
Process
// Process
//1. Input two listNodes (l1, l2)
//2. Iterate till both are not a null
// 2.1. Compare and put smaller one in resultListNode
//3. Iterate till l1 is not a null
// 3.1. Put l1 to resultListNode
//4. Iterate till l1 is not a null
// 4.1. Put l2 to resultListNode
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* tempNode = 0;
if (l1 == 0 && l2 == 0)
return tempNode;
if (l1 != 0 && l2 != 0) {
if (l1->val < l2->val) {
tempNode = new ListNode(l1->val);
l1 = l1->next;
} else {
tempNode = new ListNode(l2->val);
l2 = l2->next;
}
} else if (l1 != 0) {
tempNode = new ListNode(l1->val);
l1 = l1->next;
} else if (l2 != 0) {
tempNode = new ListNode(l2->val);
l2 = l2->next;
}
ListNode* resultListNode = tempNode;
while (l1 != 0 && l2 != 0) {
if (l1->val < l2->val) {
tempNode->next = new ListNode(l1->val);
l1 = l1->next;
} else {
tempNode->next = new ListNode(l2->val);
l2 = l2->next;
}
tempNode = tempNode->next;
}
while (l1 != 0) {
tempNode->next = new ListNode(l1->val);
tempNode = tempNode->next;
l1 = l1->next;
}
while (l2 != 0) {
tempNode->next = new ListNode(l2->val);
tempNode = tempNode->next;
l2 = l2->next;
}
return resultListNode;
}
};
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