LeetCode 657. Robot Return to Origin (C++)

题目：

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.

分析：

这道题很简单，有一个机器人站在(0，0)的位置，有四个动作分别是R (right)， L (left)， U (up)， and D (down)。在经过一系列动作后是否在(0，0)的位置。

可以定义两个值来代表垂直方向和水平方向的值，初始值是(0，0)，R就水平方向+1，L则相反，U和D同理。最后判断两个方向的值是否都为0即可。

程序：

class Solution {

public:

    bool judgeCircle(string moves) {

        int res[] = {};

        for (int i = ; i < moves.length(); i++){

            if (moves[i] == 'U')

                res[] += ;

            if (moves[i] == 'D')

                res[] -= ;

            if (moves[i] == 'R')

                res[] += ;

            if (moves[i] == 'L')

                res[] -= ;

        }

        if ((res[] == ) && (res[] == ))

            return true;

        else

            return false;

    }

};

巴特西

LeetCode 657. Robot Return to Origin (C++)

题目：

分析：

程序：

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