题目描述

Given a string path, where path[i] = 'N', 'S', 'E' or 'W', each representing moving one unit north, south, east, or west, respectively. You start at the origin (0, 0) on a 2D plane and walk on the path specified by path.

Return True if the path crosses itself at any point, that is, if at any time you are on a location you’ve previously visited. Return False otherwise.

样例

Example 1:

Input: path = "NES"
Output: false 
Explanation: Notice that the path doesn't cross any point more than once.

Example 2:

Input: path = "NESWW"
Output: true
Explanation: Notice that the path visits the origin twice.

Constraints:

• 1 <= path.length <= 10^4
• path will only consist of characters in {'N', 'S', 'E', 'W}

算法

(字符串哈希) $O(n)$

一种很直接的想法就是建立一个哈希表存储横纵坐标，每次判断当前位置是否已经被访问过了，这样就需要存储横纵坐标两个数据，比较占据存储空间。优化的办法就是只存储一个数据，但是这个数据要能反应横纵坐标的信息，所以想到把横纵坐标看成字符串，采用字符串哈希(自然溢出)的办法。

C++ 代码

class Solution {
    typedef unsigned long long ull;

    static constexpr ull Base = 13331;
    unordered_set<ull> us;
public:
    bool isPathCrossing(string path) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int x = 0, y = 0;
        insert(x, y);

        for (auto & e : path) {
            if (e == 'N') { if (!insert(x, ++y)) return true; }
            else if (e == 'S') { if (!insert(x, --y)) return true; }
            else if (e == 'E') { if (!insert(++x, y)) return true; }
            else { if (!insert(--x, y)) return true; }
        }

        return false;
    }

    bool insert(int x, int y)
    {
        string s = to_string(x) + "," + to_string(y);
        ull tmp = 0;
        for (auto & e : s) tmp = tmp * Base + e;

        if (us.find(tmp) != us.end()) return false;

        us.emplace(tmp);
        return true;
    }
};