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
b. 6th line cross with 1st line
we have covered these cases.
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 }