365天挑战LeetCode1000题——Day 111 归并排序 II_leetcode的归并排序多少题-优快云博客

本文链接：https://blog.youkuaiyun.com/ShowMeTheCod3/article/details/127279100

本文提供两种算法的实现：一种是对链表进行排序，采用归并排序的方法，通过不断拆分链表再合并的方式达到排序目的；另一种是计算数组中每个元素右侧比其小的元素个数，同样运用了归并排序的思想。

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148. 排序链表

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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* l1, ListNode* l2) {
        ListNode* pl1 = l1, *pl2 = l2;
        ListNode* ans = new ListNode();
        ListNode* curNode = ans;
        while (pl1 && pl2) {
            ListNode* tmp;
            if (pl1->val <= pl2->val) {
                tmp = new ListNode(pl1->val);
                pl1 = pl1->next;
            }
            else {
                tmp = new ListNode(pl2->val);
                pl2 = pl2->next;
            }
            curNode->next = tmp;
            curNode = tmp;
        }
        if (pl1) {
            curNode->next = pl1;
        }
        else {
            curNode->next = pl2;
        }
        return ans->next;
    }
public:
    ListNode* sortList(ListNode* head) {
        if (!head) return head;
        ListNode* dummyHead = new ListNode(-1, head);
        int length = 0;
        ListNode* cur = head;
        while (cur) {
            length++;
            cur = cur->next;
        }
        for (int curLength = 1; curLength < length; curLength *= 2) {
            ListNode* curr = dummyHead;
            ListNode* prev = dummyHead;
            while (curr) {
                ListNode* head1 = curr->next;
                for (int i = 0; i < curLength && curr; i++) {
                    curr = curr->next;
                }
                if (!curr || !curr->next) break;
                ListNode* head2 = curr->next;
                curr->next = nullptr;
                curr = new ListNode(-1, head2);
                for (int i = 0; i < curLength && curr; i++) {
                    curr = curr->next;
                }
                if (!curr || !curr->next) {
                    prev->next = mergeTwoLists(head1, head2);
                    break;
                }
                else {
                    ListNode* tmp = curr->next;
                    curr->next = nullptr;
                    prev->next = mergeTwoLists(head1, head2);
                    while (prev->next) prev = prev->next;
                    prev->next = tmp;
                    curr = prev;
                }
            }
        }
        return dummyHead->next;
    }
};

315. 计算右侧小于当前元素的个数

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代码实现

class Solution {
private:
    vector<int> ans;
    void mergeSort(vector<pair<int, int>>& tokens, vector<pair<int, int>>& tmp,
    int l, int r) {
        // cout << l << " " << r << endl;
        if (l == r) return;
        int m = (l + r) >> 1;
        mergeSort(tokens, tmp, l, m);
        mergeSort(tokens, tmp, m + 1, r);
        int i = l, j = m + 1, pos = l;
        while (i <= m && j <= r) {
            if (tokens[i].first <= tokens[j].first) {
                tmp[pos++] = tokens[i];
                ans[tokens[i].second] += j - m - 1;
                i++;
            }
            else {
                tmp[pos++] = tokens[j];
                j++;
            }
        }
        while (i <= m) {
            tmp[pos++] = tokens[i];
            ans[tokens[i].second] += j - m - 1;
            i++;
        }
        // cout << "w" << endl;
        while (j <= r) {
            tmp[pos++] = tokens[j];
            j++;
        }
        copy(tmp.begin() + l, tmp.begin() + r + 1, tokens.begin() + l);
    }
public:
    vector<int> countSmaller(vector<int>& nums) {
        int n = nums.size();
        vector<pair<int, int>> tokens;
        for (int i = 0; i < n; i++) {
            tokens.push_back({nums[i], i});
        }
        vector<pair<int, int>> tmp(n);
        ans = vector<int>(n);
        mergeSort(tokens, tmp, 0, n - 1);
        return ans;
    }
};