Self Crossing

lc335

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.

任何情况都能最后化为3种题目列举的情况

public class Solution {
public boolean isSelfCrossing(int[] x) {
    for(int i=3, l=x.length; i<l; i++) {
        if(x[i]>=x[i-2] && x[i-1]<=x[i-3]) return true;  // Case 1: current line crosses the line 3 steps ahead of it
        else if(i>=4 && x[i-1]==x[i-3] && x[i]+x[i-4]>=x[i-2]) return true; // Case 2: current line crosses the line 4 steps ahead of it
        else if(i>=5 && 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;  // Case 3: current line crosses the line 6 steps ahead of it
    }
    return false;
}
}