335. Self Crossing

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:

┌───┐
│   │
└───┼──>
    │

Input: [2,1,1,2]
Output: true

Example 2:

┌──────┐
│      │
│
│
└────────────>

Input: [1,2,3,4]
Output: false 

Example 3:

┌───┐
│   │
└───┼>

Input: [1,1,1,1]
Output: true 

 

Approach #1: Math. [Java]

class Solution {
    public boolean isSelfCrossing(int[] x) {
        if (x.length < 4) return false;
        for (int i = 3; i < x.length; ++i) {
            if (x[i] >= x[i-2] && x[i-3] >= x[i-1]) return true;
            if (i >= 4) {
                if (x[i-3] == x[i-1] && x[i] + x[i-4] >= x[i-2]) return true;
            }
            if (i >= 5) {
                if (x[i-2] >= x[i-4] && x[i-1] + x[i-5] >= x[i-3] && x[i] >= x[i-2] - x[i-4] && x[i-1] <= x[i-3]) 
                    return true;
            }
        }
        return false;
    }
}

　　

Analysis:

Categarize the self-crossing scenarios, there are 3 of them:

1. Fourth line crosses first line and works for fifth line crosses second and so on...

2. Fifth line meets first line and works for the lines after

3. Sixth line crosses first line and works for the lines after.

 

Reference:

https://leetcode.com/problems/self-crossing/discuss/79131/Java-Oms-with-explanation

 

转载于:https://www.cnblogs.com/ruruozhenhao/p/10809104.html