数据结构之多个链表问题

在数据结构中链表成了我们常见的一种形式，链表的内容学习起来比较绕，学者容易和C语言中的数组或者指针进行混淆，下面列出几道关于链表的习题供大家参考。

编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前 。

源代码

public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Partition {
    public ListNode partition(ListNode pHead, int x) {
        // < x
        ListNode less = null;
        ListNode lessLast = null;
        
        // >= x
        ListNode great = null;
        ListNode greatLast = null;
        
        ListNode cur = pHead;
        while (cur != null) {
            if (cur.val < x) {
                if (less == null) {
                    less = cur;
                } else {
                    lessLast.next = cur;
                }
                lessLast = cur;
            } else {
                if (great == null) {
                    great = cur;
                } else {
                    greatLast.next = cur;
                }
                greatLast = cur;
            }
                
            cur = cur.next;
        }
        
        if (less == null) {
            return great;
        } else {
            lessLast.next = great;
            if (greatLast != null) {
                greatLast.next = null;
            }
            
            return less;
        }
    }
}

2.

给定一个带有头结点 head 的非空单链表，返回链表的中间结点。如果有两个中间结点，则返回第二个

中间结点。

class Solution {
    private int getLength(ListNode head) {
        int len = 0;
        for (ListNode cur = head; cur != null; cur = cur.next) {
            len++;
        }
        
        return len;
    }
    public ListNode middleNode(ListNode head) {
        int len = getLength(head);
        int midLen = len / 2;
        ListNode node = head;
        for (int i = 0; i < midLen; i++) {
            node = node.next;
        }
        
        return node;
    }
}

class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        
        while (fast != null) {
            fast = fast.next;
            if (fast == null) {
                break;
            }
            slow = slow.next;
            fast = fast.next;
        }
        
        return slow;
    }
}

3.输入一个链表，输出该链表中倒数第k个结点。

public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        int len = 0;
        for (ListNode c = head; c != null; c = c.next) {
            len++;
        }
        
        if (len < k) {
            return null;
        }
        
        int steps = len - k;
        ListNode r = head;
        for (int i = 0; i < steps; i++) {
            r = r.next;
        }
        
        return r;
    }
}

public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        ListNode front = head;
        ListNode back = head;
        
        for (int i = 0; i < k; i++) {
            if (front == null) {
                return null;
            }
            front = front.next;
        }
        
        while (front != null) {
            back = back.next;
            front = front.next;
        }
        
        return back;
    }
}

4.把两个链表进行合并，并让他们进行有序排列。

public class PalindromeList {
    public ListNode getMid(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        
        while (fast != null) {
            fast = fast.next;
            if (fast == null) {
                break;
            }
            fast = fast.next;
            slow = slow.next;
        }
        
        return slow;
    }
    
    public ListNode reverse(ListNode head) {
        ListNode result = null;
        ListNode cur = head;
        while (cur != null) {
            ListNode next = cur.next;
            
            cur.next = result;
            result = cur;
            
            cur = next;
        }
        
        return result;
    }
    
    public boolean chkPalindrome(ListNode A) {
        ListNode mid = getMid(A);
        ListNode h2 = reverse(mid);
        ListNode n1 = A;
        ListNode n2 = h2;
        while (n1 != null && n2 != null) {
            if (n1.val != n2.val) {
                return false;
            }
            
            n1 = n1.next;
            n2 = n2.next;
        }
        
        return true;
    }
}

5.删除指定的所有结点

public class Solution {
    public ListNode deleteDuplication(ListNode pHead)
    {
		if (pHead == null) {
			return null;
		}
		
		ListNode prev = null;
        ListNode p1 = pHead;
        ListNode p2 = pHead.next;
        
        while (p2 != null) {
            if (p1.val != p2.val) {
                prev = p1;
                p1 = p2;
                p2 = p2.next;
            } else {
                while (p2 != null && p2.val == p1.val) {
                    p2 = p2.next;
                }
                
                if (prev == null) {
                    pHead = p2;
                } else {
                    prev.next = p2;
                }
                p1 = p2;
                if (p2 != null) {
                    p2 = p2.next;
                }
            }
        }
            
        return pHead;
    }
}

希望会对大家有帮助，有什么错误问题，还望大家指正。