題目
There is a robot starting at position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0) after it completes its moves.
The move sequence is represented by a string, and the character moves[i] represents its ith move. Valid moves are R (right), L (left), U (up), and D (down). If the robot returns to the origin after it finishes all of its moves, return true. Otherwise, return false.
Note: The way that the robot is “facing” is irrelevant. “R” will always make the robot move to the right once, “L” will always make it move left, etc. Also, assume that the magnitude of the robot’s movement is the same for each move.
Example 1:
Input: “UD”
Output: true
Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true.
Example 2:
Input: “LL”
Output: false
Explanation: The robot moves left twice. It ends up two “moves” to the left of the origin. We return false because it is not at the origin at the end of its moves.
分析
題意：
機器人站在原點上上下下左左右右，判斷它最後是否還在原點。
算法：
初始值爲（0,0）上加下減右加左減最後判斷是否還是（0,0）即可
解答
class Solution {
public boolean judgeCircle(String moves) {
int x=0;
int y=0;
for(int i=0;i<moves.length();i++){
char tmp=moves.charAt(i);
if(tmp=='U'){
y++;
}else if(tmp=='D'){
y--;
}else if(tmp=='L'){
x--;
}else{
x++;
}
}
if(x==0&&y==0){
return true;
}
return false;
}
}
表現不佳，看看別人的解法。
public class Solution {
public boolean judgeCircle(String moves) {
int x = 0;
int y = 0;
for (char ch : moves.toCharArray()) {
if (ch == 'U') y++;
else if (ch == 'D') y--;
else if (ch == 'R') x++;
else if (ch == 'L') x--;
}
return x == 0 && y == 0;
}
}
實驗證明，轉換成charArray速度更快
