【LeetCode】657. Judge Route Circle

Shihira 发布于2019-08-15 12:06 / 2311人阅读

摘要：题目描述解决方案解题思路设置初始坐标为根据上下左右指示调整坐标判断最后坐标的位置是否为起始位置。加强版循环使用比判断快方法计算向左和向右的次数是否相同，计算向上和向下的次数相同。若都相同，则回到原地。

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.

Example 1:

Input: "UD"
Output: true

Example 2:

Input: "LL"
Output: false

解决方案 - C++

class Solution {
public:
    bool judgeCircle(string moves) {
        int x = 0, y = 0;
        for(int i = 0; i < moves.length(); i++){
            if(moves.substr(i, 1) == "R"){
                x--;
            }
            if(moves.substr(i, 1) == "L"){
                x++;
            }
            if(moves.substr(i, 1) == "U"){
                y++;
            }
            if(moves.substr(i, 1) == "D"){
                y--;
            }
        }
        if(x == 0 && y == 0){
            return true;
        }else{
            return false;
        }
    } 
};

解题思路
设置初始坐标为(0,0),根据上U下D左L右R指示调整坐标,判断最后坐标的位置是否为起始位置。

Submission Details

62 / 62 test cases passed.
Status: Accepted
Runtime: 39 ms

[C++] [Java] Clean Code

Tips：

1.加强版for循环
2.使用switch比if判断快
3.Java toCharArray()方法

C++

class Solution {
public:
    bool judgeCircle(string moves) {
        int v = 0;
        int h = 0;
        for (char ch : moves) {
            switch (ch) {
                case "U" : v++; break;
                case "D" : v--; break;
                case "R" : h++; break;
                case "L" : h--; break;
            }
        }
        return v == 0 && h == 0;
    }
};

Submission Details

62 / 62 test cases passed.
Status: Accepted
Runtime: 19 ms

Java

class Solution{
    public boolean judgeCircle(String moves) {
        int v = 0, h = 0;
        for (char move : moves.toCharArray()) {
            switch (move) {
                case "U": v++; break;
                case "D": v--; break;
                case "R": h++; break;
                case "L": h--; break;
            }
        }
        return v == 0 && h == 0;
    } 
}

Submission Details

62 / 62 test cases passed.
Status: Accepted
Runtime: 11 ms

Python one liner

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

Submission Details

62 / 62 test cases passed.
Status: Accepted
Runtime: 42 ms

计算向左和向右的次数是否相同，计算向上和向下的次数相同。若都相同，则回到原地。

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