题解 | #单链表的排序#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
/*
思路：将链表复制到数组中，根据复杂度要求，对数组使用归并排序
*/

#include <array>
#include <iostream>
class Solution {
  public:
    void Merge(std::array<int, 100000>& inArr, std::array<int, 100000>& outArr,
               int start, int mid, int end) {
        int i = start;
        int j = mid + 1;
        int k = start;
        while (i <= mid || j <= end) {
            if (i > mid) {
                outArr[k++] = inArr[j++];
            } else if (j > end) {
                outArr[k++] = inArr[i++];
            } else if (inArr[i] <= inArr[j]) {
                outArr[k++] = inArr[i++];
            } else {
                outArr[k++] = inArr[j++];
            }
        }
        // important
        for (int s = start; s <= end; s++) {
            inArr[s] = outArr[s];
        }

    }
    void MergeSort(std::array<int, 100000>& inArr, std::array<int, 100000>& outArr,
                   int start, int end) {
        if (start < end) {
            auto mid = (start + end) / 2;
            MergeSort(inArr, outArr, start, mid);
            MergeSort(inArr, outArr, mid + 1, end);
            Merge(inArr, outArr, start, mid, end);
        }
    }
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类 the head node
     * @return ListNode类
     */
    ListNode* sortInList(ListNode* head) {
        // write code here
        auto tmpNode = head;
        std::array<int, 100000> inArr;
        std::array<int, 100000> outArr;
        int i = 0;
        while (tmpNode != nullptr) {
            inArr[i++] = tmpNode->val;
            tmpNode = tmpNode->next;
        }
        MergeSort(inArr, outArr, 0, i - 1);
        for (int s = 0; s < i; s++)
        {
            std::cout << inArr[s];
        }
        tmpNode = head;
        i = 0;
        while (tmpNode != nullptr) {
            tmpNode->val = outArr[i++];
            tmpNode = tmpNode->next;
        }
        return head;
    }
};