Given a callable function f(x, y) with a hidden formula and a value z, reverse engineer the formula and return *all positive integer pairs x and y where *f(x,y) == z. You may return the pairs in any order.
While the exact formula is hidden, the function is monotonically increasing, i.e.:
f(x, y) < f(x + 1, y)f(x, y) < f(x, y + 1)
The function interface is defined like this:
interface CustomFunction {
public:
// Returns some positive integer f(x, y) for two positive integers x and y based on a formula.
int f(int x, int y);
};
We will judge your solution as follows:
- The judge has a list of
9hidden implementations ofCustomFunction, along with a way to generate an answer key of all valid pairs for a specificz. - The judge will receive two inputs: a
function_id(to determine which implementation to test your code with), and the targetz. - The judge will call your
findSolutionand compare your results with the answer key. - If your results match the answer key, your solution will be
Accepted.
Example 1:
Input: function_id = 1, z = 5
Output: [[1,4],[2,3],[3,2],[4,1]]
Explanation: The hidden formula for function_id = 1 is f(x, y) = x + y.
The following positive integer values of x and y make f(x, y) equal to 5:
x=1, y=4 -> f(1, 4) = 1 + 4 = 5.
x=2, y=3 -> f(2, 3) = 2 + 3 = 5.
x=3, y=2 -> f(3, 2) = 3 + 2 = 5.
x=4, y=1 -> f(4, 1) = 4 + 1 = 5.
Example 2:
Input: function_id = 2, z = 5
Output: [[1,5],[5,1]]
Explanation: The hidden formula for function_id = 2 is f(x, y) = x * y.
The following positive integer values of x and y make f(x, y) equal to 5:
x=1, y=5 -> f(1, 5) = 1 * 5 = 5.
x=5, y=1 -> f(5, 1) = 5 * 1 = 5.
Constraints:
1 <= function_id <= 91 <= z <= 100- It is guaranteed that the solutions of
f(x, y) == zwill be in the range1 <= x, y <= 1000. - It is also guaranteed that
f(x, y)will fit in 32 bit signed integer if1 <= x, y <= 1000.
这道题说是给了一个隐藏的函数，并且给了一个函数调用的接口，可以传参数x和y进去并得到一个返回值，现在题目给定了一个返回值z，让反向找出所有的x和y的参数组合。博主刚看到这题的时候是一脸懵逼的，这尼玛还逆向工程呢，咋可能就凭一个返回值就能推导出函数表达式，瞬间尼克杨问号脸。看了看这道题的赞踩比，估计有不少人跟博主抱有同样的想法吧～ 没办法，只好逛论坛求解法，果然排在第一的仍然是 lee215 大神，他将问题做了个神转化，瞬间变的清晰无比。其实这道题并不需要知道具体的函数表达式，题目中有个关键的信息千万不能忽略，函数在x和y参数上都是单调递增的，就是说x参数不变的话，y越大，返回值越大，同理，y不变的话，x越大，返回值也越大，当然，若x和y同时大，则返回值更大。为啥说这个条件重要呢？题目中给了x和y的取值范围 [1, 1000]，则x和y可以看作一个二维数组的行列坐标，并且这个数组的值是按行和列分别递增，题目就变成了找出所有值为z的位置坐标，这样一转换，是不是豁然开朗了。
当然，暴力的方法是尝试x和y的所有的组合，一个一个的调用 API 比较返回值，但这种方法想必不能通过 OJ，毕竟完全没有考察到什么知识点。我们希望尽可能的减少调用 API 的次数，那么选择检测的起点最好是一个中间的值，数组左上角的起点是最小的数字，右下角是最大数字，那么选中间的数字就可以选左下角或右上角的数字，这里李哥选了右上角的数字当作起点，即 (1, 1000) 这个位置，然后往左下角开始遍历。对每个x和y值调用 API，若结果大于给定z值，说明需要减小参数，则y值自减1；若返回结果小于z值，说明需要增大参数，则x值自增1；若返回结果正好等于z值，完美，则将x和y值组成对儿加入结果 res 中，并且同时让x自增1，y自减1，这样最终就可以将所有和z相等的位置坐标找出来了。题目中有个标签是二分搜索法，其实就像李哥说的，如果选对了双指针的遍历逻辑，速度并不比二分搜索慢，参见代码如下：
解法一：
class Solution {
public:
vector<vector<int>> findSolution(CustomFunction& customfunction, int z) {
vector<vector<int>> res;
int x = 1, y = 1000;
while (x <= 1000 && y > 0) {
int val = customfunction.f(x, y);
if (val > z) --y;
else if (val < z) ++x;
else res.push_back({x++, y--});
}
return res;
}
};
还可以换一种写法，写得更简洁一些，但是貌似上面的解法更快一些，参见代码如下：
解法二：
class Solution {
public:
vector<vector<int>> findSolution(CustomFunction& customfunction, int z) {
vector<vector<int>> res;
int y = 1000;
for (int x = 1; x <= 1000; ++x) {
while (y > 1 && customfunction.f(x, y) > z) --y;
if (customfunction.f(x, y) == z) res.push_back({x, y});
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1237
参考资料：
https://leetcode.com/problems/find-positive-integer-solution-for-a-given-equation/
