Leetcode Self Crossing

Leetcode Self Crossing，本题的关键是找出区域的几种状态，以及一些边界情况。

区域的状态有

1. 区域越来越大，在状态下我们只有保证有x[index] > x[index - 2]即可。
2. 区域越来越小，在此状态下，需要维护一个最大值与最小值，在一般情况大有x[index] < x[index - 2]即可，但是存在一些边界情况需要了解。

边界状态有

1. 在index == 3时，进入越来越小的状态时，如: 2, 2, 3, 2，区域范围计算。
2. 在index > 3时，进入越来越小的状态时，区域范围计算。

相关代码如下：

#include<iostream>#include<vector>using namespace std;/** * The basic method of this problem is divide the state into two kind. * One is that the square is bigger and bigger. * One is that the square is smaller and smaller. * * In the first condition, we just make sure the x[index] > x[index - 2] can * fullfil this constraint. * In the second condition, we need keep the max size of the square, which need * to calculate the square size each step. */class Solution {public:    bool isSelfCrossing(vector<int>& x) {        if (x.size() <= 3) {            return false;        }        int state = 0;        int max1 = 0;        int max2 = 0;        if (x[2] <= x[0]) {            state = 1;            max1 = x[1];            max2 = x[2];        }        for (int idx = 3; idx < x.size(); idx++) {            // The first condition, the square bigger and bigger            if (state == 0) {                if (x[idx] <= x[idx - 2]) {                    state = 1;                    // state transform calculate the square size                    if ((idx == 3 && x[idx] == x[idx - 2]) ||                            (idx >= 4 && x[idx] >= x[idx - 2] - x[idx - 4])) {                        max1 = x[idx - 1] - x[idx - 3];                    } else {                        max1 = x[idx - 1];                    }                    max2 = x[idx];                }            } else {                // calculate the square size                if (x[idx] < max1) {                    int tmp = max2;                    max2 = x[idx];                    max1 = tmp;                } else {                    return true;                }            }        }        return false;    }};int main(int argc, char* argv[]) {    Solution so;    vector<int> test;    for (int i = 1; i < argc; i++) {        test.push_back(atoi(argv[i]));    }    cout<<"result: "<<so.isSelfCrossing(test)<<endl;    return 0;}

测试：./a.out 1 1 2 2 3 3 4 4 10 4 4 3 3 2 2 1 1result: 0