做科大讯飞的时候，我真是脑子瓦特了，点到第三题的时候，提交运行，结果把编程题子卷全部提交了，前面两道题还没看，敲，虽然很是无能狂怒，不过还是好好看看这些题吧，这些题是我线下思考了一下写的，没有线上测试过，不能保证AC，有发现问题的同学请提出宝贵的意见:blush:

编程题1

直接暴力循环判断吧

public class Solution {
    public int calcE(String str) {
        int res = 0, temp=0;
        char[] strs = str.toCharArray();
        for (int i = 0; i < strs.length; i++) {
            if (strs[i]=='e' || strs[i] == 'E')
                temp = 1;
            if (strs[i] = ' ' && temp == 1) {
                temp = 0;
                res++;
            }
        }
        return res;
    }
}

不保证AC

编程题2

将四边形分为两个三角形，然后计算两个三角形面积之和，三角形面积计算通过海伦公式

海伦公式，假设三角形边长分别是a，b，c，则有
$$
s = \frac{a+b+c}{2} \
S_\Delta = \sqrt{s(s-a)(s-b)(s-c)}
$$
最后将double转换为int

public class Solution {
    public int calcArea(int[] point1, int[] point2, int[] point3, int[] point4) {
        double temp =  calcTrangle(point1, point2, point3) + calcTrangle(point2, point3, point4);
        return (int)Math.ceil(temp);
    }

    public double calcTrangle(int[] point1, int[] point2, int[] point3) {
        double a = calcDistance(point1, point2);
        double b = calcDistance(point2, point3);
        double c= calcDistance(point1, point3);
        double s = (a+b+c)/2;
        return Math.sqrt(s*(s-a) * (s-b) * (s-c));

    }

    public double calcDistance(int[] point1, int[] point2) {
        return Math.sqrt((point1[0] - point2[0]) * (point1[0] - point2[0]) + (point1[1] - point2[1]) * (point1[1] - point2[1]));
    }