[Leetcode 1041] Robot Bounded In Circle

原题说明

On an infinite plane, a robot initially stands at (0, 0) and faces north.  The robot can receive one of three instructions:

• “G”: go straight 1 unit;
• “L”: turn 90 degrees to the left;
• “R”: turn 90 degress to the right.

The robot performs the instructions given in order, and repeats them forever.

Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.

 

Example 1:

Input: “GGLLGG”
Output: true
Explanation: 
The robot moves from (0,0) to (0,2), turns 180 degrees, and then returns to (0,0).
When repeating these instructions, the robot remains in the circle of radius 2 centered at the origin.

Example 2:

Input: “GG”
Output: false
Explanation: 
The robot moves north indefinitely.

Example 3:

Input: “GL”
Output: true
Explanation: 
The robot moves from (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> …

 

Note:

1. 1 <= instructions.length <= 100
2. instructions[i] is in {‘G’, ‘L’, ‘R’}

解题思路

最多执行几遍系列指令能判断是否转圈？
实际上只需要执行一遍系列指令，其导致转圈的充分必要条件是：

1. 执行完一遍指令后回到原点
   或者
2. 执行完一遍指令后发生转向（因为只有向左向右的转向，所以转向最终只可能是90，180，270）

通过变量x, y表示距离原点的位移，通过direction表示朝向，同时也是单位位移数组的index：
direction的值 0, 1, 2, 3 分别表示 北，西，南，东
对应的单位位移数组move为：{0, 1}, {-1, 0}, {0, -1}, {1, 0}
遍历instructions，对于每一个指令，改变对应的x，y和direction
最终只需要判断是否x，y均为0，或者direction不为0即可

示例代码 (cpp)

class Solution {
public:
    bool isRobotBounded(string instructions) {
        // Corresponding to N, W, S, E
        vector<vector<int>> move = {{0, 1}, {-1, 0}, {0, -1}, {1, 0}};
        int x = 0, y = 0, direction = 0;
        // char, int, float basic type doesn't need reference
        for (const auto instruction : instructions) {
            if (instruction == 'G') {
                x += move[direction][0];
                y += move[direction][1];
            } else if (instruction == 'L') {
                direction = (direction + 1) % 4;
            } else {
                // cpp could not use -1, otherwise direction < 0
                // cannot be used as move index
                direction = (direction + 3) % 4;
            }
        }
        return (x == 0 && y == 0) | (direction != 0);
    }
};

示例代码 (java)

class Solution {
    public boolean isRobotBounded(String ins) {
        int x = 0, y = 0, i = 0, d[][] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}};
        for (int j = 0; j < ins.length(); ++j)
            if (ins.charAt(j) == 'R')
                i = (i + 1) % 4;
            else if (ins.charAt(j) == 'L')
                i = (i + 3) % 4;
            else {
                x += d[i][0]; y += d[i][1];
            }
        return x == 0 && y == 0 || i > 0;
    }
}

示例代码 (python)

class Solution:
    def isRobotBounded(self, instructions: str) -> bool:
        x, y, dx, dy = 0, 0, 0, 1
        for i in instructions:
            if i == 'R': dx, dy = dy, -dx
            if i == 'L': dx, dy = -dy, dx
            if i == 'G': x, y = x + dx, y + dy
        return (x, y) == (0, 0) or (dx, dy) != (0,1)

复杂度分析

时间复杂度: O(N)
空间复杂度: O(1)

归纳总结

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