【LeetCode】Self Crossing（335）

1. 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.

　　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)


2. Answer

public class Solution {
    public boolean isSelfCrossing(int[] x) {
        // Check for initial four values manually.
        if (x.length < 4) {
            for (int el : x) {
                if (el == 0)
                    return true;
            }
            return false;
        }

        for (int i = 3; i < x.length; i++) {
            int cur = x[i];
            if (cur == 0)
                return true;
            // At any point of time, i-1 has to be less than i-3 in order to
            // intersect. Draw few figures to realize this.
            if (x[i - 1] <= x[i - 3]) {
                // Basic case. Straight forward intersection.
                //            ___
                //           |___|__  
                //               |
                //
                if (cur >= x[i - 2]) {
                    return true;
                }
                // Special case.
                if (i >= 5) {
                    // if i-2 edge is less than i-4 th edge then it cannot
                    // intersect no matter what if i < i-2 th edge.
                    //            ____
                    //           | _  | 
                    //           |__| |
                    //                |
                    if (x[i - 2] < x[i - 4])
                        continue;
                    // the intersecting case.
                    //                ____
                    //             ___|   |
                    //            |   |   |
                    //            |   |   | 
                    //            |_______|
                    //
                    if ((x[i] + x[i - 4] >= x[i - 2])
                            && (x[i - 1] + x[i - 5] >= x[i - 3]))
                        return true;
                }
            }
            // equals case
            //                 ___
            //                |   |
            //                |___|
            //
            if (i >= 4)
                if (x[i - 1] == x[i - 3] && cur + x[i - 4] == x[i - 2])
                    return true;

        }
        return false;
    }
}