题解 | #链表内指定区间反转#

/*
 * function ListNode(x){
 *   this.val = x;
 *   this.next = null;
 * }
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param head ListNode类
 * @param m int整型
 * @param n int整型
 * @return ListNode类
 */
function reverseBetween(head, m, n) {
    // write code here
    if (!head || !head.next || m === n) {
        return head;
    }
    let headKeep;
    let tailKeep;
    let between;
    let cur = head;
    let i = 1;
    while (cur && i <= n + 1) {
        let temp = cur;
        cur = cur.next;
        if (i < m) {
            if (!headKeep) {
                headKeep = temp;
                headKeep.next = null;
            } else {
                headKeep.next = temp;
                headKeep = headKeep.next;
            }
        }
        if (i >= m && i <= n) {
            if (!between) {
                between = temp;
                between.next = null;
            } else {
                temp.next = between;
                between = temp;
            }
        }
        if (i > n) {
            tailKeep = temp;
        }
        i++;
    }
    if (headKeep) {
        headKeep.next = between;
    } else {
        head = between;
    }
    if (tailKeep) {
        let btw = between;
        while (btw) {
            if (!btw.next) {
                btw.next = tailKeep;
                return head;
            }
            btw = btw.next;
        }
    }
    return head;
}
module.exports = {
    reverseBetween: reverseBetween,
};

结题思路：把链表拆分成3部分，头部不变序段，中间m-n反转段以及末尾段，中间反转，然后首中尾相连，重新组合