1237 Find Positive Integer Solution for a Given Equation 找出給定方程的正整數(shù)解

Description:
Given a function f(x, y) and a value z, return all positive integer pairs x and y where f(x,y) == z.

The function is constantly 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 positive integer f(x, y) for any given positive integer x and y.
int f(int x, int y);
};
For custom testing purposes you're given an integer function_id and a target z as input, where function_id represent one function from an secret internal list, on the examples you'll know only two functions from the list.

You may return the solutions in any order.

Example:

Example 1:

Input: function_id = 1, z = 5
Output: [[1,4],[2,3],[3,2],[4,1]]
Explanation: function_id = 1 means that f(x, y) = x + y

Example 2:

Input: function_id = 2, z = 5
Output: [[1,5],[5,1]]
Explanation: function_id = 2 means that f(x, y) = x * y

Constraints:

1 <= function_id <= 9
1 <= z <= 100
It's guaranteed that the solutions of f(x, y) == z will be on the range 1 <= x, y <= 1000
It's also guaranteed that f(x, y) will fit in 32 bit signed integer if 1 <= x, y <= 1000

題目描述:
給出一個函數(shù) f(x, y) 和一個目標結果 z，請你計算方程 f(x,y) == z 所有可能的正整數(shù) 數(shù)對 x 和 y。

給定函數(shù)是嚴格單調(diào)的，也就是說：

f(x, y) < f(x + 1, y)
f(x, y) < f(x, y + 1)
函數(shù)接口定義如下：

interface CustomFunction {
public:
// Returns positive integer f(x, y) for any given positive integer x and y.
int f(int x, int y);
};
如果你想自定義測試，你可以輸入整數(shù) function_id 和一個目標結果 z 作為輸入，其中 function_id 表示一個隱藏函數(shù)列表中的一個函數(shù)編號，題目只會告訴你列表中的 2 個函數(shù)。

你可以將滿足條件的 結果數(shù)對 按任意順序返回。

示例 :

示例 1：

輸入：function_id = 1, z = 5
輸出：[[1,4],[2,3],[3,2],[4,1]]
解釋：function_id = 1 表示 f(x, y) = x + y

示例 2：

輸入：function_id = 2, z = 5
輸出：[[1,5],[5,1]]
解釋：function_id = 2 表示 f(x, y) = x * y

提示：

1 <= function_id <= 9
1 <= z <= 100
題目保證 f(x, y) == z 的解處于 1 <= x, y <= 1000 的范圍內(nèi)。
在 1 <= x, y <= 1000 的前提下，題目保證 f(x, y) 是一個 32 位有符號整數(shù)。

思路:

注意到函數(shù)是嚴格單調(diào)的, 而且 x和 y都在 [1, 1000]的范圍內(nèi), 可以使用雙指針搜索 z的值
時間復雜度O(n), 空間復雜度O(1)

代碼:
C++:

/*
 * // This is the custom function interface.
 * // You should not implement it, or speculate about its implementation
 * class CustomFunction {
 * public:
 *     // Returns f(x, y) for any given positive integers x and y.
 *     // Note that f(x, y) is increasing with respect to both x and y.
 *     // i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
 *     int f(int x, int y);
 * };
 */

class Solution 
{
public:
    vector<vector<int>> findSolution(CustomFunction& customfunction, int z) 
    {
        vector<vector<int>> result;
        int left = 1, right = 1000;
        while (left < 1001 && right > 0) 
        {
            int temp = customfunction.f(left, right);
            if (temp == z) result.push_back({left, right});
            if (temp > z) --right;
            else ++left;
        }
        return result;
    }
};

Java:

/*
 * // This is the custom function interface.
 * // You should not implement it, or speculate about its implementation
 * class CustomFunction {
 *     // Returns f(x, y) for any given positive integers x and y.
 *     // Note that f(x, y) is increasing with respect to both x and y.
 *     // i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
 *     public int f(int x, int y);
 * };
 */
class Solution {
    public List<List<Integer>> findSolution(CustomFunction customfunction, int z) {
        List<List<Integer>> result = new ArrayList<>();
        int left = 1, right = 1000;
        while (left < 1001 && right > 0) {
            int temp = customfunction.f(left, right);
            if (temp == z) result.add(Arrays.asList(left, right));
            if (temp > z) right--;
            else left++;
        }
        return result;
    }
}

Python:

"""
   This is the custom function interface.
   You should not implement it, or speculate about its implementation
   class CustomFunction:
       # Returns f(x, y) for any given positive integers x and y.
       # Note that f(x, y) is increasing with respect to both x and y.
       # i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
       def f(self, x, y):
  
"""

class Solution:
    def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
        return filter(lambda x: customfunction.f(*x) == z, itertools.product(range(1, z + 1), repeat=2))