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]

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|       |
|       |
------------------>
        |
Input: true 
Explanation: self crossing

Example 2:

Input: [1,2,3,4]

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|       |
|
|
--------------->

Output: false 
Explanation: not self crossing

Example 3:

Input: [1,1,1,1]

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|       |
|       |
-------->

Output: true 
Explanation: self crossing

There will be a self crossing problem when the following situations happens.

All the situations’s directions to roate are counter-clockwise.

Situation 1:

Situation 2:

Situation 3:

Then we code arrcording to our above analysis.

bool isSelfCrossing(vector<int>& x)
{
    int size = x.size();
    if(size <= 3)return false;
    for(int i = 3; i < size; i++)
    {
        if(x[i] >= x[i -2] && x[i - 1] <= x[i - 3])return true;
        if(i >= 4 && x[i - 1] == x[i - 3] && x[i - 2] <= x[i] + x[i - 4])return true;
        if(i >= 5 && x[i - 3] >= x[i - 1] && x[i - 2] >= x[i - 4] && x[i - 2] <= x[i - 4] + x[i] && x[i - 3] <= x[i - 5] + x[i -1])return true;
    }   
    return false;
}
int main()
{
    return 0;
}

The third ”if” in above code includes x[i - 3] >= x[i - 1] and x[i - 2] >= x[i - 4], which is in case of the following situations.