Leetcode: Self Crossing

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)      

4th line may cross with 1st line, and so on: 5th with 2nd, ...etc

5th line may cross with 1st line, and so on: 6th with 2nd, ...etc

6th line also may cross with 1st line, and so on: 7th with 2nd, ...etc

However, if 7th line also cross with 1st line, either of the following cases should definitely happens: 

　　a. 7th line cross with 2nd line

1 public class Solution {
 2     public boolean isSelfCrossing(int[] x) {
 3         if (x.length <= 3) return false;
 4         for (int i=3; i<x.length; i++) {
 5             //check if 4th line cross with the first line and so on
 6             if (x[i]>=x[i-2] && x[i-1]<=x[i-3]) return true;
 7             
 8             //check if 5th line cross with the first line and so on
 9             if (i >= 4) {
10                 if (x[i-1]==x[i-3] && x[i]+x[i-4]>=x[i-2]) return true;
11             }
12             
13             //check if 6th line cross with the first line and so on
14             if (i >= 5) {
15                 if (x[i-2]>=x[i-4] && x[i]>=x[i-2]-x[i-4] && x[i-1]<=x[i-3] && x[i-1]>=x[i-3]-x[i-5]) return true;
16             }
17         }
18         return false;
19     }
20 }