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1237. Find Positive Integer Solution for a Given Equation - 找出给定方程的正整数解

给出一个函数  f(x, y) 和一个目标结果 z,请你计算方程 f(x,y) == z 所有可能的正整数 数对 xy

给定函数是严格单调的,也就是说:

  • f(x, y) < f(x + 1, y)
  • f(x, y) < f(x, y + 1)

函数接口定义如下:

interface CustomFunction {
public:
  // Returns positive integer f(x, y) for any given positive integer x and y.
  int f(int x, int y);
};

如果你想自定义测试,你可以输入整数 function_id 和一个目标结果 z 作为输入,其中 function_id 表示一个隐藏函数列表中的一个函数编号,题目只会告诉你列表中的 2 个函数。  

你可以将满足条件的 结果数对 按任意顺序返回。

 

示例 1:

输入:function_id = 1, z = 5
输出:[[1,4],[2,3],[3,2],[4,1]]
解释:function_id = 1 表示 f(x, y) = x + y

示例 2:

输入:function_id = 2, z = 5
输出:[[1,5],[5,1]]
解释:function_id = 2 表示 f(x, y) = x * y

 

提示:

  • 1 <= function_id <= 9
  • 1 <= z <= 100
  • 题目保证 f(x, y) == z 的解处于 1 <= x, y <= 1000 的范围内。
  • 1 <= x, y <= 1000 的前提下,题目保证 f(x, y) 是一个 32 位有符号整数。

题目标签:Math / Binary Search

题目链接:LeetCode / LeetCode中国

题解

Language Runtime Memory
cpp 0 ms 8.5 MB
/*
 * // 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>> res;
        for (int x = 1; x <= 1000; x++) {
            int left = 1, right = 1000;
            while (left <= right) {
                int mid = left + right >> 1;
                if (customfunction.f(x, mid) == z) {
                    vector<int> t;
                    t.push_back(x);
                    t.push_back(mid);
                    res.push_back(t);
                    break;
                }
                else if (customfunction.f(x, mid) > z) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
        }
        return res;
    }
};