1. 題意
2. 思路：由于是單調(diào)遞增函數(shù)，我們可以從開(kāi)始，表示在橫坐標(biāo)為以及縱坐標(biāo)為的舉行范圍內(nèi)搜索答案。分類討論：

• 如果，那么對(duì)于任意的,都有，這說(shuō)明固定，無(wú)論怎樣枚舉，都無(wú)法找到答案，那么將加一，縮小搜索范圍；
• 如果，那對(duì)于任意，都有，這說(shuō)明固定枚舉其余無(wú)法找到答案，那么將減一，縮小搜索范圍；
• 如果，那么記錄答案，同情況1一樣，將加一。由于，根據(jù)情況2，可以同時(shí)將減一。
• 不斷循環(huán)直到或為止，此時(shí)搜索范圍為空。

3. 代碼：
   C++代碼：

/*
 * // 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& f, int z) {
        vector<vector<int>> res;
        int i = 1, j = 1000;
        while(i <= 1000 && j >= 1) {
            if (f.f(i, j) == z) res.push_back({i ++, j --});
            if (f.f(i, j) > z) --j;
            if (f.f(i, j) < z) ++i;
        }
        return res;

        
    }
};

Python代碼

"""
   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: 'f', z: int) -> List[List[int]]:
        res = []
        i, j = 1, 1000
        while(i <=1000 and j >= 1):
            if customfunction.f(i, j) == z:
                res.append([i, j])
                i += 1
                j += 1
            if customfunction.f(i, j) < z:   i += 1
            if customfunction.f(i, j) > z:   j -= 1
        return res

4. 分析

• 時(shí)間復(fù)雜度：；
• 空間復(fù)雜第：。