657. Judge Route Circle-Difficulty:Easy

最新推荐文章于 2022-02-15 17:15:25 发布

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本文介绍了一种简单的方法来判断给定一系列移动指令后，一个初始位置在(0,0)的机器人是否能回到起点。通过计算上下左右移动次数是否相等来实现这一目的，并提供了C语言和Python两种实现方式。

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.

Example 1:

Input: “UD”
Output: true

Example 2:

Input: “LL”
Output: false

算法分析：只要往上走的距离等于往下走的距离，往左走的距离等于往右走的距离，即能够回到原点。

C语言版

bool judgeCircle(char* moves) {
    int r = 0, l = 0, u = 0, d = 0, i;
    for(i = 0; moves[i] != '\0'; i++)
    {
        switch(moves[i])
        {
            case 'R' : r++; break;
            case 'L' : l++; break;
            case 'U' : u++; break;
            default : d++; break;
        }
    }
    if(r == l && u == d)
        return true;
    else
        return false;
}

Python版

class Solution(object):
    def judgeCircle(self, moves):
        """
        :type moves: str
        :rtype: bool
        """
        #一行入魂
        return moves.count('R') == moves.count('L') and moves.count('U') == moves.count('D')