Algorithm

[LeetCode 335] Self Crossing [Java] [Runtime : 0MS]

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2017-07-13

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07/13

1. Runtime Distribution

2. Submission Details

3. 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.

4. Code

public boolean isSelfCrossing(int[] x){
    int length = x.length;

    if(length < 4){
        return false;
    }

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

5.Test

import org.junit.Test;

public class LeetCode0335 {

    public boolean isSelfCrossing(int[] x){
        int length = x.length;

        if(length < 4){
            return false;
        }

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

    @Test
    public void test(){
        int [] x = {2, 1, 1, 2};
        System.out.println(isSelfCrossing(x));
    }
}

Math

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