hcx2013
- 浏览: 88861 次
-
最新评论
-
hcx2013：
mircle 写道Given nums = [1, 2, 1, ...
Single Number III -
mircle：
Given nums = [1, 2, 1, 3, 2, 5， ...
Single Number III -
hcx2013：
恩，可以，代码更简洁
空格替换 -
QuarterLifeForJava：
这样可以吗？
public static void main( ...
空格替换 -
hcx2013：
这个2是指多出2位，按你的理解可以这么改：int len = ...
空格替换
文章列表
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
...
Balanced Binary Tree
- 博客分类：
- leetcode
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
/**
* Definition for a binary tree node.
* public class TreeNode {
* i ...
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
/**
* Definition for a b ...
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class S ...
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
...
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.set(key, value) - Set or insert the value if the key ...
Given inorder and postorder traversal of a tree, construct the binary tree.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public Tre ...
Given preorder and inorder traversal of a tree, construct the binary tree.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ...
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
...
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its zigzag level order ...
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
/* ...
Symmetric Tree
- 博客分类：
- leetcode
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
/**
* Definition for a binary tre ...
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?
confused what "{1,#,2,3}" means? > read more on how binary t ...
Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
...
题目描述
利用字符重复出现的次数，编写一个方法，实现基本的字符串压缩功能。比如，字符串“aabcccccaaa”经压缩会变成“a2b1c5a3”。若压缩后的字符串没有变短，则返回原先的字符串。
给定一个string iniString为待压缩的串(长度小于等于3000)，保证串内字符均由大小写英文字母组成，返回一个string，为所求的压缩后或未变化的串。
测试样例
"aabcccccaaa"
返回："a2b1c5a3"
"welcometonowcoderrrrr"
返回："welcometonow ...