Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each move is represent by a characte…

原题链接在这里:https://leetcode.com/problems/judge-route-circle/description/ 题目: Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move seq…

Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each move is represent by a characte…

题目描述 初始位置 (0, 0) 处有一个机器人.给出它的一系列动作,判断这个机器人的移动路线是否形成一个圆圈,换言之就是判断它是否会移回到原来的位置. 移动顺序由一个字符串表示.每一个动作都是由一个字符来表示的.机器人有效的动作有 R(右),L(左),U(上)和 D(下).输出应为 true 或 false,表示机器人移动路线是否成圈. 示例 1: 输入: "UD" 输出: true 示例 2: 输入: "LL" 输出: false 思路 设置初始点的坐标为x=0…

[LeetCode]657. Judge Route Circle 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/judge-route-circle/description/ 题目描述: Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which mea…

657. Judge Route Circle[easy] Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each m…

Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each move is represent by a characte…

［抄题］: Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each move is represent by a ch…

Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each move is represent by a characte…

static int wing=[]() { std::ios::sync_with_stdio(false); cin.tie(NULL); ; }(); class Solution { public: bool judgeCircle(string moves) { ,count2=; for(char c:moves) { if(c=='U') count1++; else if(c=='D') count1--; else if(c=='L') count2++; else count…

Given an integer (signed 32 bits), write a function to check whether it is a power of 4. Example: Given num = 16, return true. Given num = 5, return false. Follow up: Could you solve it without loops/recursion? Credits:Special thanks to @yukuairoy fo…

Given an integer, write a function to determine if it is a power of three. Follow up:Could you do it without using any loop / recursion? Credits:Special thanks to @dietpepsi for adding this problem and creating all test cases. 这道题让我们判断一个数是不是3的次方数,在Le…

Given an integer, write a function to determine if it is a power of two. Hint: Could you solve it in O(1) time and using O(1) space? 这道题让我们判断一个数是否为2的次方数,而且要求时间和空间复杂度都为常数,那么对于这种玩数字的题,我们应该首先考虑位操作 Bit Operation.在LeetCode中,位操作的题有很多,比如比如Repeated DNA Seque…

In the video game Fallout 4, the quest "Road to Freedom" requires players to reach a metal dial called the "Freedom Trail Ring", and use the dial to spell a specific keyword in order to open the door. Given a string ring, which represe…

  思路: 主要判断左子树与右子树. 在判断左时,循环下去肯定会到达叶子结点中最左边的结点与最右边的结点比较. 到了这一步因为他们都没有左(右)子树了,所以得开始判断这两个结点的右(左)子树了. 当某个结点对称了,它的左子树也对称了,右子树也对称了,那才是真的对称了. C++ /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode…

Determine whether an integer is a palindrome. Do this without extra space.(不要使用额外的空间) Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using extra space. You…

题意:如题 思路:递归解决,同判断对称树的原理差不多.先保证当前两个结点是相等的,再递归保证两左结点是相等的,再递归保证右结点是相等的. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Sol…

题目: Determine whether an integer is a palindrome. Do this without extra space. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using extra space. You could…

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following [1,2,2,null,3,null,3] is not: / \ \ \ Note:Bonus po…

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. Note: For the purpose of this problem, we define empty string as valid palindrome. Example 1: Input: "A man, a plan, a canal: Panama" O…

一.问题描述 判断一个integer 型的数字是否是回文,空间复杂度应该是常数级别的 . 二.问题分析 首先,负数不是回文,10的整数倍不会是回文,个位数一定是回文. 三.代码实现 思路:将一个数字翻转,即最高位变成最低位,最低位变成最高位,然后比较输入的字符和翻转之后的字符. class Solution { bool isPalindrome(int x) { || (x% == && x != )) { return false; } int i = x; // 将 x 保存起来 ;…

Question: Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Analysis: 只想到了,首先判断是否有环,若有环,则总链首开始,一次判断是否是环的开始,这样T(O) = O(n^2). 其实是一个数学问题,详细思路参照链接http://blog.csdn.net/sbitswc/article/details/27584037 . Answer…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 判断链表中是否有环,不能用额外的空间,可以使用快慢指针,慢指针一次走一步,快指针一次走两步,若是有环则快慢指针会相遇,若是fast->next==NULL则没有环. 值得注意的是:在链表的题中,快慢指针的使用频率还是很高,值得注意. /** * Definition for si…

dfs遍历一下判断 public boolean isSameTree(TreeNode p, TreeNode q) { if (p==null) { return q == null; } else { if (q==null || p.val!=q.val) return false; else { return (isSameTree(p.left,q.left)&&isSameTree(p.right,q.right)); } } }…

/* 思路是判断26个字符在两个字符串中出现的次数是不是都一样,如果一样就返回true. 记住这个方法 */ if (s.length()!=t.length()) return false; int[] words = new int[26]; for (int i = 0; i < s.length(); i++) { words[s.charAt(i)-'a']++; words[t.charAt(i)-'a']--; } for (int i = 0; i < 26; i++) { i…

问题: 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.   分析: 考虑使用深度优先遍历的方法,同时遍历两棵树,遇到不等的就返回. 代码如下: /** * Definition f…

题意: 如果所给序列的元素不是唯一的,则返回true,否则false. 思路: 哈希map解决. class Solution { public: bool containsDuplicate(vector<int>& nums) { unordered_map<int,int> mapp; ; i<nums.size(); i++) { if(mapp[nums[i]]) return true; ; } return false; } }; AC代码 python…

题意:如题,平衡树是指任意一个节点(除了叶子),其左子树的高度与右子树的高度相差不超过1. 思路:递归解决,但是提供的函数不满足递归的要求啊,我们至少得知道高度,又得返回真假,所以另开个函数解决. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), ri…

Given two binary trees, write a function to check if they are the same or not. Two binary trees are considered the same if they are structurally identical and the nodes have the same value. Example 1: Input: / \ / \ [,,], [,,] Output: true Example 2:…

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree [1,2,2,3,4,4,3] is symmetric:    1   / \  2   2 / \   / \3  4 4  3 But the following [1,2,2,null,3,null,3] is not:    1   / \ …