657. Judge Route Circle【easy】】的更多相关文章

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…
[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…
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…
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…
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…
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