LeetCode 21 Merge Two Sorted Lists
2017-01-30 15:15
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题意:
排序两个已经有序的链表,空间复杂度O(1)。
思路:
很基本的题,经典的2个指针遍历链表的思路,将小的那个元素接到答案的最后一个元素之后,直到一条链遍历完毕。
代码:
/**
* 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) {
if (l1 == NULL) {
return l2;
}
if (l2 == NULL) {
return l1;
}
ListNode *ans, *head;
if (l1->val <= l2->val) {
ans = head = l1;
l1 = l1->next;
} else {
ans = head = l2;
l2 = l2->next;
}
while (l1 != NULL && l2 != NULL) {
if (l1->val <= l2->val) {
head->next = l1;
l1 = l1->next;
} else {
head->next = l2;
l2 = l2->next;
}
head = head->next;
}
if (l1 != NULL && l2 == NULL) {
head->next = l1;
} else if (l1 == NULL && l2 != NULL) {
head->next = l2;
}
return ans;
}
};
排序两个已经有序的链表,空间复杂度O(1)。
思路:
很基本的题,经典的2个指针遍历链表的思路,将小的那个元素接到答案的最后一个元素之后,直到一条链遍历完毕。
代码:
/**
* 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) {
if (l1 == NULL) {
return l2;
}
if (l2 == NULL) {
return l1;
}
ListNode *ans, *head;
if (l1->val <= l2->val) {
ans = head = l1;
l1 = l1->next;
} else {
ans = head = l2;
l2 = l2->next;
}
while (l1 != NULL && l2 != NULL) {
if (l1->val <= l2->val) {
head->next = l1;
l1 = l1->next;
} else {
head->next = l2;
l2 = l2->next;
}
head = head->next;
}
if (l1 != NULL && l2 == NULL) {
head->next = l1;
} else if (l1 == NULL && l2 != NULL) {
head->next = l2;
}
return ans;
}
};
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