[LeetCode]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:

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)

注意走的方向是一定的，所以如果没有环的话是往外扩展的回型圈，或者是向内的。

如果出现环可能有三种情况如下。

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