LeetCode Self Crossing

Description:

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.

Solution:

總結一下，就三種可能發生cross的情況。

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