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:

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)

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

按顺时针画线，问是否会出现相交。对于某一条边，出现相交的情况有三种，如图所示：

根据上面三种情况编写就可以了。

代码：

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