题目一
链表题——反转链表
根据单链表的头节点head来返回反转后的链表
具体题目如下
解法
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/** * 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 reverseList(ListNode head) { ListNode pre,cur,nxt; pre = null ; cur = head; nxt = head; while (cur!= null ){ nxt = cur.next; cur.next = pre; pre = cur; cur = nxt; } return pre; } } |
题目二
链表题——反转链表
按照一定数量的节点来进行反转并返回反转之后的链表
具体题目如下
解法
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/** * 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 reverseKGroup(ListNode head, int k) { if (head == null ) return null ; ListNode a, b; a = b = head; for ( int i = 0 ; i < k; i++) { if (b == null ) return head; b = b.next; } ListNode newHead = reverse(a, b); a.next = reverseKGroup(b, k); return newHead; } ListNode reverse(ListNode a, ListNode b) { ListNode pre,cur,nxt; pre = null ; cur = a; nxt = a; while (cur!=b){ nxt = cur.next; cur.next = pre; pre = cur; cur = nxt; } return pre; } } |
题目三
链表题——回文链表
根据单链表的头节点head来判断该链表是否是回文链表,并返回结果
具体题目如下
解法:后序遍历与left比较
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/** * 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 { ListNode left; public boolean isPalindrome(ListNode head) { left = head; return traverse(head); } boolean traverse(ListNode right){ if (right == null ) return true ; boolean res = traverse(right.next); res = res && (right.val == left.val); left = left.next; return res; } } |
题目四
二叉树题——翻转二叉树
根据所给的二叉树根节点root来翻转此二叉树,并返回翻转后的二叉树根节点
具体题目如下
解法
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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public TreeNode invertTree(TreeNode root) { if (root== null ){ return null ; } TreeNode lf = invertTree(root.left); TreeNode rg = invertTree(root.right); root.left = rg; root.right = lf; return root; } } |
题目五
二叉树题——填充节点
给定一个完美二叉树,填充该二叉树每个节点的下一个右侧节点指针
具体题目如下
解法
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/* // Definition for a Node. class Node { public int val; public Node left; public Node right; public Node next; public Node() {} public Node(int _val) { val = _val; } public Node(int _val, Node _left, Node _right, Node _next) { val = _val; left = _left; right = _right; next = _next; } }; */ class Solution { public Node connect(Node root) { if (root== null ) return null ; method(root.left,root.right); return root; } public void method(Node left,Node right){ if (left == null || right == null ) { return ; } left.next = right; method(left.left,left.right); method(right.left,right.right); method(left.right,right.left); } } |
题目六
二叉树链表题——将二叉树展开为链表
根据给定的二叉树根节点root,将此二叉树展开为单链表
具体题目如下
解法
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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public void flatten(TreeNode root) { if (root == null ) return ; flatten(root.left); flatten(root.right); TreeNode left = root.left; TreeNode right = root.right; root.left = null ; root.right = left; TreeNode p = root; while (p.right != null ) { p = p.right; } p.right = right; } } |
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原文链接:https://blog.csdn.net/wai_58934/article/details/123031314