leetcode 725. Split Linked List in Parts 链表均匀分割

比眉伴天荒 2022-06-03 08:58 100阅读 0赞

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 \[1, 50\].

本题题意看起来有点复杂，其实还是很简单的，就是分个链表，要求尽量分割均匀，而且相差不能超过1个，同时链表前面的组的元素数量不可以少于后面的，那么可以这么做，首先求一个平均数k，和余数ext，给前ext个组多一个元素（也即k+1个元素），后面的都按照k个去分配，那么直接遍历即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    /*
    struct ListNode 
    {
         int val;
         ListNode *next;
         ListNode(int x) : val(x), next(NULL) {}
    };
    */
    
    class Solution 
    {
    public:
        vector<ListNode*> splitListToParts(ListNode* root, int k) 
        {
            vector<ListNode*> res(k,NULL);
            ListNode* tmp = root;
            int len = 0;
            while (tmp != NULL)
            {
                len++;
                tmp = tmp->next;
            }
    
            int aver = len / k, ext = len % k;
            for (int i = 0; i < k && root!=NULL; i++)
            {
                res[i] = root;
                int subsize = aver + (i < ext ? 1 : 0);
                for (int j = 1; j < subsize; j++)
                    root = root->next;
    
                ListNode* tt = root->next;
                root->next = NULL;
                root = tt;
            }
            return res;
        }
    };