【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 = [code][2, 1, 1, 2]

,
┌───┐
│ │
└───┼──>
│Return true (self crossing)
[/code]
　　Example 2:

Given x = [code][1, 2, 3, 4]

,
┌──────┐
│ │
│
│
└────────────>Return false (not self crossing)
[/code]
　　Example 3:

Given x = [code][1, 1, 1, 1]

,
┌───┐
│ │
└───┼>Return true (self crossing)2. Answer[/code]

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;
}
}