[数据结构与算法]725. 分隔链表

?

链接：725. 分隔链表

题解：力扣

?

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    vector<ListNode*> splitListToParts(ListNode* head, int k) {
        vector<ListNode*> res(k, nullptr);
        if (!head) {
            return res;
        }
        int len = 0;
        ListNode* cur = head;
        while (cur) {
            ++len;
            cur = cur->next;
        }
        int cnt = len/k; // 每段几个
        int remain = len%k; // 补充的余数
        int i = 0;
        ListNode* prev = head;
        while (head) {
            res[i++] = head; // 每一段的开始
            for (int i = 0; i < cnt; ++i) {
                if (head) {
                    prev = head;
                    head = head->next;
                }
            }
            if (remain > 0) {
                prev = head;
                head = head->next;
                --remain;
            }
            // 结束一段
            prev->next = nullptr;
        }
        // RVO
        return res;
    }
};

?