[Leetcode]657. Robot Return to Origin

Easy

There 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.

题目大意：一个机器人根据给定的字符串的指令进行移动（没有移动方向导致的机器人面朝影响，即：如同人朝东南西北移动一样）U是向上移动，D是向下移动，L是向左移动，R是向右移动。如果移动结束后机器人又回到原位则返回true，否则返回false。

可以用一个数组来记录机器人相对于原点的移动，移动结束后如果数组中的数据是0就表示机器人相对原点移动是0，也就是没有移动，返回true，否则返回false。那么这个数组就要有两项，一项用于记录左右移动结果，一项用于记录上下移动结果。代码如下：

class Solution {
public:
    bool judgeCircle(string moves) {
        if(moves.size()==0)return true;
        vector<int> step(2,0);
        for(char c: moves){
            if(c==‘L‘){
                step[0]++;
            }
            else if(c==‘R‘){
                step[0]--;
            }
            else if(c==‘U‘){
                step[1]++;
            }
            else if(c==‘D‘){
                step[1]--;
            }
        }
        if(step[0]==0 && step[1]==0){
            return true;
        }
        else return false;
    }
};

原文地址：https://www.cnblogs.com/cff2121/p/11421053.html