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)

这题考验逻辑归类能力。看了别人的思路，其实只有三种情况能crossing

如果在一个轮回内，情况比较简单，只有一种可能crossing。

如果超过了一个轮回，有两种情况。

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