leetcode 657. Judge Route Circle

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


解法1：

class Solution(object):
    def judgeCircle(self, moves):
        """
        :type moves: str
        :rtype: bool
        out("")=True
        out("L" or "D" or "U" or "R") =False
        out("LR")=True
        out("RL")=True
        out("RLDU")=True
        out("LLR")=False
        """
        x = y = 0
        for m in moves:
            if m == "L":
                x -= 1
            elif m == "R":
                x += 1
            elif m == "U":
                y += 1
            else:
                y -= 1
        return x==0 and y==0

用查找表更好：

class Solution(object):
    def judgeCircle(self, moves):
        """
        :type moves: str
        :rtype: bool
        """
        x,y = 0,0
        offsets = {"U":[0,1], "D":[0,-1], "R":[1,0], "L":[-1,0]}
        for move in moves:
            x,y = x+offsets[move][0], y+offsets[move][1]
        return (x == 0) and (y == 0)

解法2：

def judgeCircle(self, moves):
    return moves.count('L') == moves.count('R') and moves.count('U') == moves.count('D')

直接统计LR数目是否相等，同时UD数目是否相等。

类似代码：

def judgeCircle(self, moves):
    c = collections.Counter(moves)
    return c['L'] == c['R'] and c['U'] == c['D']

因为：

>>> import collections
>>> collections.Counter("abca")
Counter({'a': 2, 'b': 1, 'c': 1})