Tag: 快慢指针

Posted 2022-05-06Updated 2022-05-05LeetCode / Easy / 复习2 minutes read (About 353 words)

Question

Write an algorithm to determine if a number n is happy.

A happy number is a number defined by the following process:

Starting with any positive integer, replace the number by the sum of the squares of its digits.

Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1.

Those numbers for which this process ends in 1 are happy.

Return true if n is a happy number, and false if not.

Solution

首先证明，快乐数的计算不会趋近于无穷大。
我们可以考虑每一个位数的最大数字（9999…999）。
其下一个数值等于(81*n)。
对于三位数以下，其最大的值得下一个结果不会大于243。
对于四位数以上，下一个结果会快速收缩到三位数。
因此计算结果不会趋近于无穷。

当我们计算快乐数的值时，形成了一个隐式链表。
因此我们可以用快慢指针的方式，查询链表上是否有环。
快指针比慢指针初始值快1个单位，且移动速度为慢指针的一倍。
如果链表中有环，则两者终能相遇。
如果fast指针先走到1，等于走到了链表的终点。

Code

class Solution {
    public boolean isHappy(int n) {
        int slow = n, fast = getNext(n);
        while(fast != 1 && fast != slow){
            fast = getNext(getNext(fast));
            slow = getNext(slow);
        }
        return fast == 1;
    }
    
    private int getNext(int n){
        int sum = 0;
        while(n != 0){
            int digit = n % 10;
            sum += (digit * digit);
            n = n / 10;
        }
        return sum;
    }
}

Posted 2022-04-25Updated 2022-04-25LeetCode / Medium / 复习3 minutes read (About 411 words)

142. Linked List Cycle II

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

快慢指针。快指针的移动速度是慢指针的两倍。
设环外长度为a，b是快指针和慢指针相遇的位置，c是环中剩余位置。
可以由此得到公式a + (n + 1)b + nc = 2(a + b)，也就是a = c + (n - 1)(b + c)
由于（b + c）是环的长度。因此，当两个指针相遇时，在头部设置一个新节点。慢指针和新指针将在循环入口处相遇，此时返回节点。

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if( head == null || head.next == null) return null;
        ListNode slow = head;
        ListNode fast = head;
        
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow.equals(fast)) break;
        }
        
        if(fast == null || fast.next == null){
            return null;
        }
        
        ListNode root = head;
        while(!root.equals(slow)){
            slow = slow.next;
            root = root.next;
        }
        return root;
    }
}

哈希表，递归并将节点加入哈希集合，如果重复则返回节点，反之返回null。

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    HashSet<ListNode> set;
    public ListNode detectCycle(ListNode head) {
        set = new HashSet<>();
        return findCycle(head, 0);
    }
    
    private ListNode findCycle(ListNode root, int count){
        if(root == null){
            return null;
        }
        if(set.contains(root)){
            return root;
        }
        set.add(root);
        count++;
        return findCycle(root.next, count);
    }
}

Posted 2022-04-09Updated 2022-04-08LeetCode / Easy / 复习a minute read (About 201 words)

141. Linked List Cycle

问题
Given head, the head of a linked list, determine if the linked list has a cycle in it.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to. Note that pos is not passed as a parameter.

Return true if there is a cycle in the linked list. Otherwise, return false.

遍历并移动快慢指针。
如两个指针最终相遇，则链表中有循环。
如快指针移动到链表尾部，则链表无循环。

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        if ( head == null){
            return false;
        }
        else if ( fast.next == null){
            return false;
        }
        
        while( fast != null && fast.next != null ){
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast){
                return true;
            }
        }
        return false;
    }
}

Posted 2022-04-07Updated 2022-04-08LeetCode / Easya minute read (About 132 words)

876. Middle of the Linked List

问题
Given the head of a singly linked list, return the middle node of the linked list.

If there are two middle nodes, return the second middle node.

快慢指针，两个指针不同速度遍历链表。当快指针达到链表尾部时候，慢指针正好在中间。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {
        
        ListNode slowNode = head;
        ListNode fastNode = head;
        
        while( fastNode != null ){
            fastNode = fastNode.next;
            if (fastNode == null){
                return slowNode;
            }
            slowNode = slowNode.next;
            fastNode = fastNode.next;
        }
        return slowNode;
    }
}

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