README_EN.md

335. Self Crossing

中文文档

Description

You are given an array of integers distance.

You start at point (0,0) on an X-Y plane and you move distance[0] meters to the north, then distance[1] meters to the west, distance[2] meters to the south, distance[3] meters to the east, and so on. In other words, after each move, your direction changes counter-clockwise.

Return true if your path crosses itself, and false if it does not.

 

Example 1:

Input: distance = [2,1,1,2]
Output: true

Example 2:

Input: distance = [1,2,3,4]
Output: false

Example 3:

Input: distance = [1,1,1,1]
Output: true

 

Constraints:

• 1 <= distance.length <= 500
• 1 <= distance[i] <= 500

 

Follow up: Could you write a one-pass algorithm with O(1) extra space?

Solutions

                i-2
    case 1 : i-1┌─┐
                └─┼─>i
                 i-3

                   i-2
    case 2 : i-1 ┌────┐
                 └─══>┘i-3
                 i  i-4

    case 3 :    i-4
               ┌──┐
               │i<┼─┐
            i-3│ i-5│i-1
               └────┘
                i-2

Python3

class Solution:
    def isSelfCrossing(self, distance: List[int]) -> bool:
        d = distance
        for i in range(3, len(d)):
            if d[i] >= d[i - 2] and d[i - 1] <= d[i - 3]:
                return True
            if i >= 4 and d[i - 1] == d[i - 3] and d[i] + d[i - 4] >= d[i - 2]:
                return True
            if i >= 5 and d[i - 2] >= d[i - 4] and d[i - 1] <= d[i - 3] and d[i] >= d[i - 2] - d[i - 4] and d[i - 1] + d[i - 5] >= d[i - 3]:
                return True
        return False

Java

class Solution {
    public boolean isSelfCrossing(int[] distance) {
        int[] d = distance;
        for (int i = 3; i < d.length; ++i) {
            if (d[i] >= d[i - 2] && d[i - 1] <= d[i - 3]) {
                return true;
            }
            if (i >= 4 && d[i - 1] == d[i - 3] && d[i] + d[i - 4] >= d[i - 2]) {
                return true;
            }
            if (i >= 5 && d[i - 2] >= d[i - 4] && d[i - 1] <= d[i - 3] && d[i] >= d[i - 2] - d[i - 4] && d[i - 1] + d[i - 5] >= d[i - 3]) {
                return true;
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool isSelfCrossing(vector<int>& distance) {
        vector<int> d = distance;
        for (int i = 3; i < d.size(); ++i)
        {
            if (d[i] >= d[i - 2] && d[i - 1] <= d[i - 3]) return true;
            if (i >= 4 && d[i - 1] == d[i - 3] && d[i] + d[i - 4] >= d[i - 2]) return true;
            if (i >= 5 && d[i - 2] >= d[i - 4] && d[i - 1] <= d[i - 3] && d[i] >= d[i - 2] - d[i - 4] && d[i - 1] + d[i - 5] >= d[i - 3]) return true;
        }
        return false;
    }
};

Go

func isSelfCrossing(distance []int) bool {
	d := distance
	for i := 3; i < len(d); i++ {
		if d[i] >= d[i-2] && d[i-1] <= d[i-3] {
			return true
		}
		if i >= 4 && d[i-1] == d[i-3] && d[i]+d[i-4] >= d[i-2] {
			return true
		}
		if i >= 5 && d[i-2] >= d[i-4] && d[i-1] <= d[i-3] && d[i] >= d[i-2]-d[i-4] && d[i-1]+d[i-5] >= d[i-3] {
			return true
		}
	}
	return false
}

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