标签:leetcode als return aci his for rip bool bsp
here is a robot starting at position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0) after it completes its moves.
The move sequence is represented by a string, and the character moves[i] represents its ith move. Valid moves are R (right), L (left), U (up), and D (down). If the robot returns to the origin after it finishes all of its moves, return true. Otherwise, return false.
Note: The way that the robot is "facing" is irrelevant. "R" will always make the robot move to the right once, "L" will always make it move left, etc. Also, assume that the magnitude of the robot‘s movement is the same for each move.
Example 1:
Input: "UD" Output: true Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true.
Example 2:
Input: "LL" Output: false Explanation: The robot moves left twice. It ends up two "moves" to the left of the origin. We return false because it is not at the origin at the end of its moves.
class Solution { public boolean judgeCircle(String moves) { if(moves ==null|| moves.length() ==0){ return true; } int Count_Up = 0; int Count_Down = 0; int Count_Left = 0; int Count_Right = 0; for(int i = 0; i<moves.length(); i++){ if(moves.charAt(i)==‘U‘){ Count_Up ++; } if(moves.charAt(i)==‘D‘){ Count_Down++; } if(moves.charAt(i)==‘L‘){ Count_Left++; } if(moves.charAt(i)==‘R‘){ Count_Right++; } } if(Count_Left!= Count_Right || Count_Up != Count_Down){ return false; } else{ return true; } } }
(Easy) Robot Return to Origin LeetCode
标签:leetcode als return aci his for rip bool bsp
原文地址:https://www.cnblogs.com/codingyangmao/p/11304656.html