88.merge-sorted-array

       这道题目被喷的很惨,而且难度极低,基本上是个人都能做出来,但是我要说的是,一个细节的处理很重要,个人感觉是能反映出一些灵活运用的能力。

       题目大体意思是这样的,给定两个从小到大的排序数组,将两个数组归并排序到第一数组里面去,第一个数组中前m个是待排元素,后n个为位置是任意数字,目的是能够放下两个数组。

      这道题用常规的归并排序完全可以做,但是,既然题目专门将一个数组的大小扩充了,意味着作者并不像让我们重新申请空间,因此,根据题目我们可以从后往前做归并。这思路不难,但是有时候对于我们这样的新手可能不是那么容易想到。

具体的代码如下:

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int k = m+n;
        --m, --n;
        while (m>=0 && n>=0) {
            if (nums1[m] > nums2[n]) {
                nums1[--k] = nums1[m--];
            } else {
                nums1[--k] = nums2[n--];
            }
        }
        while (m>=0) {
            nums1[--k] = nums1[m--];
        }
        while (n>=0) {
            nums1[--k] = nums2[n--];
        }
    }
};

 

To merge k sorted linked lists, one approach is to repeatedly merge two of the linked lists until all k lists have been merged into one. We can use a priority queue to keep track of the minimum element across all k linked lists at any given time. Here's the code to implement this idea: ``` struct ListNode { int val; ListNode* next; ListNode(int x) : val(x), next(NULL) {} }; // Custom comparator for the priority queue struct CompareNode { bool operator()(const ListNode* node1, const ListNode* node2) const { return node1->val > node2->val; } }; ListNode* mergeKLists(vector<ListNode*>& lists) { priority_queue<ListNode*, vector<ListNode*>, CompareNode> pq; for (ListNode* list : lists) { if (list) { pq.push(list); } } ListNode* dummy = new ListNode(-1); ListNode* curr = dummy; while (!pq.empty()) { ListNode* node = pq.top(); pq.pop(); curr->next = node; curr = curr->next; if (node->next) { pq.push(node->next); } } return dummy->next; } ``` We start by initializing a priority queue with all the head nodes of the k linked lists. We use a custom comparator that compares the values of two nodes and returns true if the first node's value is less than the second node's value. We then create a dummy node to serve as the head of the merged linked list, and a current node to keep track of the last node in the merged linked list. We repeatedly pop the minimum node from the priority queue and append it to the merged linked list. If the popped node has a next node, we push it onto the priority queue. Once the priority queue is empty, we return the head of the merged linked list. Note that this implementation has a time complexity of O(n log k), where n is the total number of nodes across all k linked lists, and a space complexity of O(k).
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