Leetcode 3200. 三角形的最大高度

原题链接：Leetcode 3200. 三角形的最大高度

模拟1：

class Solution {
public:
    int maxHeightOfTriangle(int red, int blue) {
        int tmp1 = 0, tmp2 = 0;
        int n1 = 0, n2 = 0;
        int ji_sum = n1 * n1;
        int ou_sum = n2 * n2 + n2;
        // 红1 蓝2
        for (int i = 0; i <= 10; i++) {
            if ((i * i) <= red && ((i + 1) * (i + 1)) > red)
                n1 = i;
        }
        for (int i = 0; i <= 9; i++) {
            if ((i * i + i) <= blue && (i * i + 3 * i + 2) > blue)
                n2 = i;
        }
        tmp1 = n2 >= n1 ? 2 * n1 : 2 * n2 + 1;
        n1 = 0;
        n2 = 0;
        // 蓝1 红2
        for (int i = 1; i <= 10; i++) {
            if ((i * i) <= blue && ((i + 1) * (i + 1)) > blue)
                n1 = i;
        }
        for (int i = 1; i <= 9; i++) {
            if ((i * i + i) <= red && (i * i + 3 * i + 2) > red)
                n2 = i;
        }
        tmp2 = n2 >= n1 ? 2 * n1 : 2 * n2 + 1;
        return max(tmp1, tmp2);
    }
};

模拟2：这种方法好理解很多

class Solution {
public:
    int maxheight(int ji, int ou) {
        for (int i = 1;; i++) {
            if (i % 2 == 1) {
                ji -= i;
                if (ji < 0)
                    return i - 1;
            } else {
                ou -= i;
                if (ou < 0)
                    return i - 1;
            }
        }
    }
    int maxHeightOfTriangle(int red, int blue) {
        return max(maxheight(red, blue), maxheight(blue, red));
    }
};