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
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.
Note:
1 <= path.length <= 10^4
path will only consist of characters in {'N', 'S', 'E', 'W}
解析
根据题意，就是如果点根据方向移动的位置又回到了以前走过的位置则返回 True ，否则返回 False 。思路简单，就是用列表将每个点的位置都记录下来，只要之后移动的位置也在列表中存在就返回 True ，否则将该位置加入到列表中，循环这个步骤，直到结束。
解答
class Solution(object):
def isPathCrossing(self, path):
"""
:type path: str
:rtype: bool
"""
points = [(0, 0)]
for c in path:
point = None
if c == 'N':
point = (points[-1][0], points[-1][1] + 1)
elif c == 'S':
point = (points[-1][0], points[-1][1] - 1)
elif c == 'E':
point = (points[-1][0] + 1, points[-1][1])
elif c == 'W':
point = (points[-1][0] - 1, points[-1][1])
if point not in points:
points.append(point)
else:
return True
return False
运行结果
Runtime: 8 ms, faster than 100.00% of Python online submissions for Path Crossing.
Memory Usage: 13.8 MB, less than 30.63% of Python online submissions for Path Crossing.
原题链接：leetcode.com/problems/pa…
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