20190730算法题存档

题目描述

Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-2→…

You must do this in-place without altering the nodes' values.

For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}.

public class Solution {public void reorderList(ListNode head) {if(head == null || head.next == null) {return;}ListNode slow = head;ListNode quick = head;while(quick.next != null && quick.next.next != null) {slow = slow.next;quick = quick.next.next;}ListNode mid = slow.next;if(mid != null && mid.next != null) {ListNode next = mid.next;while(next != null) {mid.next = next.next;next.next = slow.next;slow.next = next;next = mid.next;}}ListNode p = head;ListNode q = slow.next;while(q != null && p != null){slow.next = q.next;q.next = p.next;p.next = q;p = p.next.next;q = slow.next;}}
}

题目描述

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

Follow up:
Can you solve it without using extra space?

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

题目描述

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

public class Solution {public boolean hasCycle(ListNode head) {if(head != null) {ListNode slow = head;ListNode fast = head;while(fast != null && fast.next != null) {fast = fast.next.next;slow = slow.next;if(fast == slow) {return true;}}}return false;}
}

题目描述

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s ="leetcode",
dict =["leet", "code"].

Return true because"leetcode"can be segmented as"leet code".

public class Solution {public boolean wordBreak(String s, Set<String> dict) {boolean[] temp = new boolean[s.length() + 1];temp[0] = true;for(int i = 1; i < temp.length; i++) {for(int j = 0; j < i; j++) {if (temp[j] && dict.contains(s.substring(j, i))){temp[i] = true;break;}}}return temp[temp.length - 1];}
}

题目描述

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

public class Solution {public RandomListNode copyRandomList(RandomListNode head) {if(head == null || (head.next == null && head.random == null)){return head;}RandomListNode node = head;while(node != null){RandomListNode copy = new RandomListNode(node.label);copy.next = node.next;copy.random = node.random;node.next = copy;node = node.next.next;}node = head.next;while(node != null){if(node.next != null){node.next = node.next.next;}if(node.random != null){node.random  = node.random.next;}node = node.next;}return head.next;}}