Leetcode: Self Crossing

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.

这题有点意思，一开始只是想到了向内旋转的情况，觉得并没有什么难度。其实要看当前的旋转是向内还是向外的。如果是向内旋转，在遍历到第i个元素的时候，只要这个元素没有超出第i-2个元素即可；如果是向外旋转，需要做一些判断，因为向外旋转是可能转换成向内旋转的（然而向内旋转不可能转换成向外旋转，这点很重要）。判断的时候要注意一下会不会和3步之前的元素有交叉，如果有，可以在原地直接修改x[i-1]的值。

代码：

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