1、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 means it moves back to the original place.
The move sequence is represented by a string. And each move is represent by a character. The valid robot moves are R (Right), L (Left), U (Up) and D (down).
The output should be true or false representing whether the robot makes a circle.
题目的意思是，如果一个可以上下左右运动，分别记做 ‘U’ ,'D'，‘L’，‘R’。将一个机器人的运动序列记做一个string ,根据这个string 判断机器人是否回到了原点。
2、问题分析
特例是机器人没有动，即string的长度是0,那么机器人就在原位。
机器人回到原位的条件是，上下运动的步数相等 并且 左右运动的步数相等。遍历一次string ，分别记录上下左右运动的次数，然后比较每个方向上运动的次数。如果 R == L && U == D ,那么机器人回到了原点。
3、代码
1 if(moves.size() == 0) 2 return true; 3 4 int u = 0; 5 int d = 0; 6 int l = 0; 7 int r = 0; 8 9 int i = 0; 10 while(i < moves.size()) 11 { 12 switch (moves[i]) 13 { 14 case 'U': 15 ++u; 16 break; 17 case 'D': 18 ++d; 19 break; 20 case 'L': 21 ++l; 22 break; 23 case 'R': 24 ++r; 25 break; 26 default: 27 break; 28 } 29 i++; 30 } 31 32 if(u == d && l == r) 33 return true; 34 35 return false;