题解 | #输入序列连续的序列检测#

public class Solution { public ListNode reverseBetween(ListNode head, int m, int n) {
//计算length，同时把mPre和nNext找到出来 ListNode tmp = head; ListNode mPre = null; ListNode nNext = null; int len = 0; for(; tmp!=null; tmp = tmp.next){ len = len+1; if(len == m -1) mPre = tmp; if(len == n + 1) nNext = tmp; } //判断1<=m <= n <= length if(m > n || m < 1 || n > len) return head;

    //先把m-n反转，再设置mPre nNext的连接
    //三个辅助指针用于反转
    ListNode node1 = (mPre == null)?head : mPre.next;//指向m
    ListNode node2 = node1.next;
    ListNode node3 = null;
    node1.next = nNext;//m的next指向n的next
    //开始反转
    while(node2 != nNext){
        node3 = node2.next;
        node2.next = node1;
        node1 = node2;
        node2 = node3;
    }//循环结束，node1指向n
    //判断是否换头，根据m是否为head
    if(mPre == null){
        return node1;
    }else{
        mPre.next = node1;
        return head;
    }
}

}