LeetCode 335. Self Crossing

最新推荐文章于 2024-08-28 07:25:21 发布

junchen1992

最新推荐文章于 2024-08-28 07:25:21 发布

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分类专栏：算法 LeetCode

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本文介绍了一种使用O(1)额外空间的一过式算法来判断由一系列正数构成的路径是否会发生自我交叉。通过改变方向并利用简洁的条件判断，该算法能有效地解决这一问题，并附带提供了具体的Python实现代码。

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Problem Statement

(Source) You are given an array x of n positive numbers. You start at point (0,0) and moves x[0] metres to the north, then x[1] metres to the west, x[2] metres to the south, x[3] metres to the east and so on. In other words, after each move your direction changes counter-clockwise.

Write a one-pass algorithm with O(1) extra space to determine, if your path crosses itself, or not.

Example 1:

Given x = 
[2, 1, 1, 2]
,
┌───┐
│   │
└───┼──>
    │

Return true (self crossing)

Example 2:

Given x = 
[1, 2, 3, 4]
,
┌──────┐
│      │
│
│
└────────────>

Return false (not self crossing)

Example 3:

Given x = 
[1, 1, 1, 1]
,
┌───┐
│   │
└───┼>

Return true (self crossing)

Tags: Math.

Solution

class Solution(object):
    def isSelfCrossing(self, x):
        """
        :type x: List[int]
        :rtype: bool
        """
        b = c = d = e = 0
        for a in x:
            if d >= b > 0 and (a >= c or a >= c-e >= 0 and f >= d-b):
                return True
            b, c, d, e, f = a, b, c, d, e
        return False