leetcode Sort List

最新推荐文章于 2022-05-05 15:09:07 发布

原创最新推荐文章于 2022-05-05 15:09:07 发布 · 200 阅读

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探讨了在O(ologn)时间复杂度下对链表进行排序的方法，对比了快速排序和归并排序在链表场景下的应用，最终采用链表归并排序实现更优效率。

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没想到什么好办法，要将O(ologn)时间里完成，我先遍历一遍将链表存储到数组里，再对数组进行排序，排序方法写了快排和归并的，过是能过，但是时间要140ms，最快的就是直接对链表进行归并排序，第一次写比较难受，用快慢指针来分成两个链表，具体思想还是和数组的归并排序一样。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

void merge(int A[],int L1,int R1,int L2,int R2){
    int i = L1, j = R2; // i指向A[L1] j指向A[L2]
    int temp[100000], Index = 0;
    while (i <= R1 && j<=R2) {
        if (A[i] <= A[j])
            temp[Index++] = A[i++];
        else
            temp[Index++] = A[j++];
    }
    while (i <= R1) temp[Index++] = A[i++];
    while (j <= R2) temp[Index++] = A[j++];
    for (int i =0; i<Index; i++)
        A[L1 + i] = temp[i];
}

void merge_sort(int A[],int left,int right){
    if (left < right){
        int mid = (left+right) / 2;
        merge_sort(A, left, mid);
        merge_sort(A, mid+1, right);
        merge(A, left, mid,mid+1,right);
    }
}
int Partition(int A[],int left,int right){
    int temp = A[left];
    while (left < right) {
        while (left < right && A[right] > temp)
            right--;
        A[left] = A[right];
        while (left <right && A[left] <= temp)
            left++;
        A[right] = A[left];
    }
    A[left] = temp;
    return left;
}

void quick_sort(int A[],int left,int right){
    if (left < right) {
        int pos = Partition(A, left, right);
        quick_sort(A, left, pos-1);
        quick_sort(A, pos+1, right);
    }
}


struct ListNode* sortList(struct ListNode* head) {
    if(!head || !head->next)
        return head;
    int data[100000];
    int index = 0; // the length of the list
    struct ListNode *current = head;
    while(current){
        data[index++] = current->val;
        current = current->next;
    }
    quick_sort(data,0,index-1);
    current = head;
    index = 0;
    while(current){
        current->val = data[index++];
        current = current->next;
    }
    return head;
}