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
这道题让我们判断一个路径是否绕圈，就是说有多少个U，就得对应多少个D。同理，L和R的个数也得相等。这不就是之前那道Valid Parentheses的变种么，这次博主终于举一反三了！这比括号那题还要简单，因为括号至少还有三种，这里就水平和竖直两种。比较简单的方法就是使用两个计数器，如果是U，则cnt1自增1；如果是D，cnt1自减1。同理，如果是L，则cnt1自增1；如果是R，cnt1自减1。最后只要看cnt1和cnt2是否同时为0即可，参见代码如下：
解法一：
class Solution {
public:
bool judgeCircle(string moves) {
int cnt1 = , cnt2 = ;
for (char move : moves) {
if (move == 'U') ++cnt1;
else if (move == 'D') --cnt1;
else if (move == 'L') ++cnt2;
else if (move == 'R') --cnt2;
}
return cnt1 == && cnt2 == ;
}
};
下面这种解法使用了哈希表来建立字符和其出现的次数之间的映射，最后直接比较对应的字符出现的次数是否相等即可，参见代码如下：
解法二：
class Solution {
public:
bool judgeCircle(string moves) {
unordered_map<char, int> m;
for (char c : moves) ++m[c];
return m['L'] == m['R'] && m['U'] == m['D'];
}
};
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参考资料：
https://discuss.leetcode.com/topic/99256/c-counter-4-lines-solution