365天挑战LeetCode1000题——Day 111 归并排序 I

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原创最新推荐文章于 2026-11-08 10:44:48 发布 189 阅读 AI 写同款· GEO 优化 ›

【信息科学与工程学】【数据科学】数据科学领域第六篇算法设计 07 堆、栈、表、矩阵、集合、队列、数组、字符串、树、图、动态规划、结构体、元组的算法01 当前段必须包含所有字符的最后出现位置。1. 字符串分割成最多片段2. 基因序列功能域划分3. 文本段落划分4. 音频声道分离5. 图像色彩分区6. 社交网络社群划分7. 蛋白质结构域8. 音乐乐章分段9. 代码模块划分10. 章节划分特征：一次扫描，贪心确定边界。阅读详情

合并两个排序的链表

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代码实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* left, ListNode* right) {
        // if one of the two is empty return
        if (!left) return right;
        if (!right) return left;
        ListNode* head = new ListNode(-1);
        ListNode* traverse = head;
        while (left && right) {
            if (left->val < right->val) {
                traverse->next = new ListNode(left->val);
                left = left->next;
            }
            else {
                traverse->next = new ListNode(right->val);
                right = right->next;
            }
            traverse = traverse->next;
            
        }
        if (left) traverse->next = left;
        if (right) traverse->next = right;
        return head->next;
    }
};

合并K个升序链表

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代码实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
private:
    ListNode* mergeTwoLists(ListNode* left, ListNode* right) {
        // if one of the two is empty return
        if (!left) return right;
        if (!right) return left;
        ListNode* head = new ListNode(-1);
        ListNode* traverse = head;
        while (left && right) {
            if (left->val < right->val) {
                traverse->next = new ListNode(left->val);
                left = left->next;
            }
            else {
                traverse->next = new ListNode(right->val);
                right = right->next;
            }
            traverse = traverse->next;
            
        }
        if (left) traverse->next = left;
        if (right) traverse->next = right;
        return head->next;
    }
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.empty()) return nullptr;
        vector<ListNode*> tmp;
        int n = -1;
        while (lists.size() > 1) {
            n = lists.size();
            tmp.clear();
            for (int i = 0; i < n; i += 2) {
                if (i < n - 1) {
                    tmp.push_back(mergeTwoLists(lists[i], lists[i + 1]));
                }
                else {
                    tmp.push_back(lists[i]);
                }
            }
            lists = tmp;
            // cout << lists.size() << endl;
        }
        return lists[0];
    }
};

剑指 Offer 51. 数组中的逆序对

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代码实现

class Solution {
private:
    int mergeSort(vector<int>& nums, vector<int>& tmp, int l, int r) {
        if (l == r) return 0;
        int m = (l + r) >> 1;
        int count = mergeSort(nums, tmp, l, m) + mergeSort(nums, tmp, m + 1, r);
        int i = l, j = m + 1, pos = l;
        while (i <= m && j <= r) {
            if (nums[i] <= nums[j]) {
                tmp[pos] = nums[i];
                i++;
                count += j - m - 1;
            }
            else {
                tmp[pos] = nums[j];
                j++;
            }
            pos++;
        }
        while (i <= m) {
            tmp[pos++] = nums[i++];
            count += j - m - 1;
        }
        while (j <= r) {
            tmp[pos++] = nums[j++];
        }
        copy(tmp.begin() + l, tmp.begin() + r + 1, nums.begin() + l);
        return count;
    }
public:
    int reversePairs(vector<int>& nums) {
        if (nums.empty()) return {};
        int n = nums.size();
        vector<int> tmp(n);
        return mergeSort(nums, tmp, 0, n - 1);
    }
};