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摘要：題目描述解決方案解題思路設(shè)置初始坐標(biāo)為根據(jù)上下左右指示調(diào)整坐標(biāo)判斷最后坐標(biāo)的位置是否為起始位置。加強(qiáng)版循環(huán)使用比判斷快方法計(jì)算向左和向右的次數(shù)是否相同，計(jì)算向上和向下的次數(shù)相同。若都相同，則回到原地。
題目描述
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解決方案 - 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;
}
}
};
解題思路
設(shè)置初始坐標(biāo)為(0,0),根據(jù)上U下D左L右R指示調(diào)整坐標(biāo),判斷最后坐標(biāo)的位置是否為起始位置。
Submission Details
62 / 62 test cases passed. Status: Accepted Runtime: 39 ms[C++] [Java] Clean Code
Tips：
1.加強(qiáng)版for循環(huán)
2.使用switch比if判斷快
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 msPython 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
計(jì)算向左和向右的次數(shù)是否相同，計(jì)算向上和向下的次數(shù)相同。若都相同，則回到原地。
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