[LeetCode] Judge Route Circle 判斷路線繞圈

 

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

 

這道題讓我們判斷一個路徑是否繞圈，就是說有多少個U，就得對應多少個D。同理，L和R的個數也得相等。這不就是之前那道Valid Parentheses的變種么，這次博主終於舉一反三了！這比括號那題還要簡單，因為括號至少還有三種，這里就水平和豎直兩種。比較簡單的方法就是使用兩個計數器，如果是U，則cnt1自增1；如果是D，cnt1自減1。同理，如果是L，則cnt1自增1；如果是R，cnt1自減1。最后只要看cnt1和cnt2是否同時為0即可，參見代碼如下：

 

解法一：

class Solution {
public:
    bool judgeCircle(string moves) {
        int cnt1 = 0, cnt2 = 0;
        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 == 0 && cnt2 == 0;
    }
};

 

下面這種解法使用了哈希表來建立字符和其出現的次數之間的映射，最后直接比較對應的字符出現的次數是否相等即可，參見代碼如下：

 

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'];
    }
};

 

類似題目：

Valid Parentheses

 

參考資料： 

https://discuss.leetcode.com/topic/99256/c-counter-4-lines-solution 

 

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