【leetcode】1237. Find Positive Integer Solution for a Given Equation

题目如下：

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 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

解题思路：看到1 <= x, y <= 1000时，就可以意识到O(n^2)的复杂度是可以接受的，那么两层循环计算一下吧。

代码如下：

"""
   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(object):
    def findSolution(self, customfunction, z):
        """
        :type num: int
        :type z: int
        :rtype: List[List[int]]
        """
        res = []
        for x in range(1,1001):
            for y in range(1,1001):
                if customfunction.f(x,y) == z:
                    res.append([x,y])
                elif customfunction.f(x,y) > z:
                    break
        return res