[LeetCode] Split Linked List in Parts

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题目描述

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].
输入是一个链表，要求把链表尽量平分为k 份，如果不能完全平分，要求划分后的每部分最多相差不超过一个且size 更大的部分处于相对前面的位置。
解题思路是遍历一遍求出链表的长度，然后求出每部分的长度以及不能平分的长度。顺序遍历链表，按照分组要求不断截断链表后插入。

C++ 实现

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/**
 * 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* x = root;
        while(x)
        {
            size++;
            x = x->next;
        }

        const int part_size = size / k;
        const int left = size % k;
        // 直接初始化大小为k的数组，这样不需要对part_size为0的情况进行特殊处理
        vector<ListNode*> res(k);
        for (int i = 0; i < k && root; i++)
        {
            res[i] = root;
            // 前面几个部分需要均摊掉left的部分，每部分消费一个
            for (int j = 1; j < part_size + (i < left); j++)
            {
                root = root->next;
            }
            ListNode* next = root->next;
            // 截断链表
            root->next = nullptr;
            // 更新游标位置
            root = next;
        }
        return res;
    }
};
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