第88天。

今天的题目是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].

虽然不难，但是自己还是做了一个早上，现在的效率真的是低的可以。

首先要知道每个块要有多少个节点就必须先知道总共有多少个节点，所以第一遍扫描算节点数肯定是少不了的，然后就可以开始算每个块的个数了。

• k>=n时，显然前n块只需要放一个就好了，后面的不用管。
• k<n时，可以分成两部分，前面的一部分比后面的一部分多放一个块，穷举几次可以知道，后面的部分每个快放n/k个节点，前面的部分有n%k个块。

其实到这里就可以发现其实并不需要分情况,第二种已经包括了第一种了。

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vector<ListNode*> splitListToParts(ListNode* root, int k) {
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    int size = 0;
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    for(ListNode *p = root;p != nullptr;p=p->next) size++;
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    vector<ListNode *> ret(k,nullptr);
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    int a = size/k;
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    int b = size%k;
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    ListNode *pre = nullptr;
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    ListNode *p = root;
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    for(int i = 0;i < k && p;i++) {
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        ret[i] = p;
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        for(int j = 0;j < a + (b>0);j++) {
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            pre = p;
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            p = p->next;
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        }
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        b--;
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        pre->next = nullptr;
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    }
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    return ret;
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}