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)

解题思路

相交可分为如下三种情况：

• 第四条线与第一条线相交
• 第五条线与第一条线相交或重叠
• 第六条线与第一条线相交
  
  点Xi为给定的最后一个点。

实现代码

// Runtime: 1 ms
public class Solution {
    public boolean isSelfCrossing(int[] x) {
        int len = x.length;
        if(len <= 3) return false;

        for(int i = 3; i < len; i++) {
            if(x[i] >= x[i-2] && x[i-1] <= x[i-3]) {
                return true;
            }
            if(i >= 4) {
                if(x[i-1] == x[i-3] && x[i] + x[i-4] >= x[i-2]) {
                    return true;
                }
            }
            if(i >= 5) {
                if(x[i-2] - x[i-4] >= 0 && x[i] >= x[i-2] - x[i-4] && x[i-1] >= x[i-3] - x[i-5] && x[i-1] <= x[i-3]) {
                    return true; 
                }
            }
        }
        return false;
    }
}