[leetcode] 335. Self Crossing 解题报告

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题目链接：https://leetcode.com/problems/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)

思路：一种比较简单的思路是判断是否图形是否一直处于螺旋增加状态或者螺旋减小状态，如果是的话那么就不会相交．螺旋增加就是对于i >= 2, 始终有x[i] > x[i-2]，螺旋减小就是对于x>=2，始终有x[i] < x[i-2]. 如果其不能一直处于这两种状态的一种，那么还可以分为：

１．从螺旋减小到到螺旋增大，这种状态必然是相交的

２．从螺旋增大到螺旋减小，这种状态并不一定相交，但是会改变x[i-1]边界条件．即我们需要判断x[i]+x[i-4]是否比x[i-2]小，如果小的话x[i-1]的边界不会变，但是如果x[i]+x[i-4] >=x[i-2]，x[i-1]将会减小为x[i-1] - x[i-3]，画个图就可以很明显看出这两种情况的区别．另外需要注意一个case就是i=3时是没有x[i-4]因此比较的时候将其当成0即可．

代码如下:

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

参考:https://leetcode.com/discuss/88196/re-post-2-o-n-c-0ms-solutions

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