[数据结构与算法]数据结构（链表相关面试题的）（6）

前言

数据结构和算法笔记

一、快慢指针

1）输入链表头节点，奇数长度返回中点，偶数长度返回上中点

2）输入链表头节点，奇数长度返回中点，偶数长度返回下中点

3）输入链表头节点，奇数长度返回中点前一个，偶数长度返回上中点前一个

4）输入链表头节点，奇数长度返回中点前一个，偶数长度返回下中点前一个

public static class Node {
		public int value;
		public Node next;

		public Node(int v) {
			value = v;
		}
	}

	// head 头
	public static Node midOrUpMidNode(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return head;
		}
		// 链表有3个点或以上
		Node slow = head.next;
		Node fast = head.next.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

	public static Node midOrDownMidNode(Node head) {
		if (head == null || head.next == null) {
			return head;
		}
		Node slow = head.next;
		Node fast = head.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

	public static Node midOrUpMidPreNode(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return null;
		}
		Node slow = head;
		Node fast = head.next.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

	public static Node midOrDownMidPreNode(Node head) {
		if (head == null || head.next == null) {
			return null;
		}
		if (head.next.next == null) {
			return head;
		}
		Node slow = head;
		Node fast = head.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

	public static Node right1(Node head) {
		if (head == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get((arr.size() - 1) / 2);
	}

	public static Node right2(Node head) {
		if (head == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get(arr.size() / 2);
	}

	public static Node right3(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get((arr.size() - 3) / 2);
	}

	public static Node right4(Node head) {
		if (head == null || head.next == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get((arr.size() - 2) / 2);
	}

二、判断回文数

1.利用栈

public static boolean isPalindrome1(Node head) {
		Stack<Node> stack = new Stack<Node>();
		Node cur = head;
		while (cur != null) {
			stack.push(cur);
			cur = cur.next;
		}
		while (head != null) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

2.省空间（省一半栈空间）

public static boolean isPalindrome2(Node head) {
		if (head == null || head.next == null) {
			return true;
		}
		Node right = head.next;
		Node cur = head;
		while (cur.next != null && cur.next.next != null) {
			right = right.next;
			cur = cur.next.next;
		}
		Stack<Node> stack = new Stack<Node>();
		while (right != null) {
			stack.push(right);
			right = right.next;
		}
		while (!stack.isEmpty()) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

3.不用额外空间

// need O(1) extra space
	public static boolean isPalindrome3(Node head) {
		if (head == null || head.next == null) {
			return true;
		}
		Node n1 = head;
		Node n2 = head;
		while (n2.next != null && n2.next.next != null) { // find mid node
			n1 = n1.next; // n1 -> mid
			n2 = n2.next.next; // n2 -> end
		}
		// n1 中点
		
		
		n2 = n1.next; // n2 -> right part first node
		n1.next = null; // mid.next -> null
		Node n3 = null;
		while (n2 != null) { // right part convert
			n3 = n2.next; // n3 -> save next node
			n2.next = n1; // next of right node convert
			n1 = n2; // n1 move
			n2 = n3; // n2 move
		}
		n3 = n1; // n3 -> save last node
		n2 = head;// n2 -> left first node
		boolean res = true;
		while (n1 != null && n2 != null) { // check palindrome
			if (n1.value != n2.value) {
				res = false;
				break;
			}
			n1 = n1.next; // left to mid
			n2 = n2.next; // right to mid
		}
		n1 = n3.next;
		n3.next = null;
		while (n1 != null) { // recover list
			n2 = n1.next;
			n1.next = n3;
			n3 = n1;
			n1 = n2;
		}
		return res;
	}

三、将单向链表按某值划分成左边小、中间相等、右边大的形式

1.把链表放入数组里，在数组上做partition（笔试用）

public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			this.value = data;
		}
	}

	public static Node listPartition1(Node head, int pivot) {
		if (head == null) {
			return head;
		}
		Node cur = head;
		int i = 0;
		while (cur != null) {
			i++;
			cur = cur.next;
		}
		Node[] nodeArr = new Node[i];
		i = 0;
		cur = head;
		for (i = 0; i != nodeArr.length; i++) {
			nodeArr[i] = cur;
			cur = cur.next;
		}
		arrPartition(nodeArr, pivot);
		for (i = 1; i != nodeArr.length; i++) {
			nodeArr[i - 1].next = nodeArr[i];
		}
		nodeArr[i - 1].next = null;
		return nodeArr[0];
	}

	public static void arrPartition(Node[] nodeArr, int pivot) {
		int small = -1;
		int big = nodeArr.length;
		int index = 0;
		while (index != big) {
			if (nodeArr[index].value < pivot) {
				swap(nodeArr, ++small, index++);
			} else if (nodeArr[index].value == pivot) {
				index++;
			} else {
				swap(nodeArr, --big, index);
			}
		}
	}

	public static void swap(Node[] nodeArr, int a, int b) {
		Node tmp = nodeArr[a];
		nodeArr[a] = nodeArr[b];
		nodeArr[b] = tmp;
	}

2.分成小、中、大三部分，再把各个部分之间串起来（面试用）

public static Node listPartition2(Node head, int pivot) {
		Node sH = null; // small head
		Node sT = null; // small tail
		Node eH = null; // equal head
		Node eT = null; // equal tail
		Node mH = null; // big head
		Node mT = null; // big tail
		Node next = null; // save next node
		// every node distributed to three lists
		while (head != null) {
			next = head.next;
			head.next = null;
			if (head.value < pivot) {
				if (sH == null) {
					sH = head;
					sT = head;
				} else {
					sT.next = head;
					sT = head;
				}
			} else if (head.value == pivot) {
				if (eH == null) {
					eH = head;
					eT = head;
				} else {
					eT.next = head;
					eT = head;
				}
			} else {
				if (mH == null) {
					mH = head;
					mT = head;
				} else {
					mT.next = head;
					mT = head;
				}
			}
			head = next;
		}
		// 小于区域的尾巴，连等于区域的头，等于区域的尾巴连大于区域的头
		if (sT != null) { // 如果有小于区域
			sT.next = eH;
			eT = eT == null ? sT : eT; // 下一步，谁去连大于区域的头，谁就变成eT
		}
		// 下一步，一定是需要用eT 去接 大于区域的头
		// 有等于区域，eT -> 等于区域的尾结点
		// 无等于区域，eT -> 小于区域的尾结点
		// eT 尽量不为空的尾巴节点
		if (eT != null) { // 如果小于区域和等于区域，不是都没有
			eT.next = mH;
		}
		return sH != null ? sH : (eH != null ? eH : mH);
	}

四、rand指针

一种特殊的单链表节点类描述如下
class Node {
int value;
Node next;
Node rand;
Node(int val) { value = val; }
}
rand指针是单链表节点结构中新增的指针，rand可能指向链表中的任意一个节点，也可能指向null。
给定一个由Node节点类型组成的无环单链表的头节点 head，请实现一个函数完成这个链表的复制，并返回复制的新链表的头节点。
【要求】
时间复杂度O(N)，额外空间复杂度O(1)

public static class Node {
		int val;
		Node next;
		Node random;

		public Node(int val) {
			this.val = val;
			this.next = null;
			this.random = null;
		}
	}

	public static Node copyRandomList1(Node head) {
		// key 老节点
		// value 新节点
		HashMap<Node, Node> map = new HashMap<Node, Node>();
		Node cur = head;
		while (cur != null) {
			map.put(cur, new Node(cur.val));
			cur = cur.next;
		}
		cur = head;
		while (cur != null) {
			// cur 老
			// map.get(cur) 新
			// 新.next ->  cur.next克隆节点找到
			map.get(cur).next = map.get(cur.next);
			map.get(cur).random = map.get(cur.random);
			cur = cur.next;
		}
		return map.get(head);
	}

五、给定两个可能有环也可能无环的单链表，头节点head1和head2。请实现一个函数，如果两个链表相交，请返回相交的 第一个节点。如果不相交，返回null

【要求】
如果两个链表长度之和为N，时间复杂度请达到O(N)，额外空间复杂度 请达到O(1)。

解决思路分为两大步：

（1）找到单链表的入环节点，无环则返回null

（2）根据得到的链表的入环节点，再展开讨论

找到单链表的入环节点方法：

/**
 * 找到单链表的入环节点，无环则返回null
 *
 */
public static Node getLoopNode(Node head) {
    if (head == null || head.next == null || head.next.next == null) {
        return null;
    }
    // 快指针
    Node fast = head.next.next;
    // 慢指针
    Node slow = head.next;
    while (fast != slow) {
        // 如果某次快指针走着走着变为空了，则此单链表必定无环
        if (fast.next == null || fast.next.next == null) {
            return null;
        }
        fast = fast.next.next;
        slow = slow.next;
    }

    // 当快慢指针相遇时，快指针再次从头节点开始走
    fast = head;
    while (fast != slow) {
        // 此时就不需要再判断为空了，能走到这里则此单链表必定是有环的
        // 此时快指针一次走一个节点
        fast = fast.next;
        // 慢指针依然一次走一个节点
        slow = slow.next;
    }

    return slow;
}

两个链表都无环：

/**
 * 现在已经知道两个链表无环，找到它们的第一个相交节点
 *
 */
public static Node noLoop(Node head1, Node head2) {
    if (head1 == null || head2 == null) {
        return null;
    }

    Node current1 = head1;
    Node current2 = head2;
    int n = 0;
    // 链表一遍历一遍得到长度、最后一个节点
    while (current1.next != null) {
        n++;
        current1 = current1.next;
    }

    // 链表二遍历一遍得到长度差、最后一个节点
    while (current2.next != null) {
        n--;
        current2 = current2.next;
    }

    // 如果两个链表的最后一个节点不等，则一定不相交
    if (current1 != current2) {
        return null;
    }

    // 长链表的头节点给current1
    current1 = n > 0 ? head1 : head2;
    // 短链表的头节点给current2
    current2 = current1 == head1 ? head2 : head1;

    n = Math.abs(n);
    // 长链表先走n个节点
    while (n != 0) {
        n--;
        current1 = current1.next;
    }

    // 两个链表挨个往下走，第一次相等的节点即是第一个相交的节点
    while (current1 != current2) {
        current1 = current1.next;
        current2 = current2.next;
    }

    return current1;
}

六、能不能不给单链表的头节点，只给想要删除的节点，就能做到在链表上把这个点删掉？

不给头节点是没有办法完成的

总结