leetcode 725. Split Linked List in Parts-优快云博客

本文介绍了一种算法，该算法能够将一个单链表均匀地分割成k个连续的部分。每部分长度尽可能相等，相差不超过1。通过示例展示了如何进行链表分割，并提供了完整的C++实现代码。

725. Split Linked List in Parts

Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts".

The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.

The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.

Return a List of ListNode's representing the linked list parts that are formed.

Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]

Example 1:

Input: root = [1, 2, 3], k = 5 Output: [[1],[2],[3],[],[]] Explanation: The input and each element of the output are ListNodes, not arrays. For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null. The first element output[0] has output[0].val = 1, output[0].next = null. The last element output[4] is null, but it's string representation as a ListNode is [].

Example 2:

Input: 
root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
Explanation:
The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.

Note:

The length of root will be in the range [0, 1000].
Each value of a node in the input will be an integer in the range [0, 999].
k will be an integer in the range [1, 50].

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    vector<ListNode*> splitListToParts(ListNode* root, int k) 
    {
        int size = 0;
        ListNode* p = root;
        while (p)
        {
            size++;
            p = p->next;
        }
        int jiben = size / k;
        int duoyu = size % k; //余数 代表 前面几项会比后面几项多一个
        
        vector<ListNode*> ret;
        ListNode* now_root = root;
        for (int i = 0; i < k; i++)   
        {
            ret.push_back(now_root);
            if (!now_root) continue;
            int flag = (duoyu-- > 0) ? 1 : 0; //多一个的项flag = 1
            for (int j = 1; j < jiben + flag; j++)
            {
                now_root = now_root->next;
            }
            ListNode* last = now_root;
            now_root = last->next;
            last->next = NULL; //断尾
        }
        return ret;
    }
    
};