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📜  最小化达到N值所需的步骤

📅  最后修改于: 2021-04-24 17:59:02             🧑  作者: Mango

给定范围为[-INFINITY，+ INFINITY]的无穷大行和整数N ，任务是通过向前移动i或向后移动1步，找到从0开始达到N所需的最小移动次数。在我的第一个举动中。

范例：

方法：最初的想法是继续添加1，2，3。 。 。 。 K ，直到大于或等于所需值N为止。然后，计算从当前总和中减去所需的数字。执行以下步骤解决该问题：

  • 最初，增加K直到N大于当前值。现在，停在某个位置
    pos = 1 + 2 +…………。 +步数=步数*(步数+ 1)/ 2≥N.
    注意： 0≤pos – N <步长。否则，最后一步是不可能的。
    • 情况1：如果pos = N，则“ steps”是必需的答案。
    • 情况2：如果pos≠N ，则用-1替换K的任何迭代。
  • 通过用-1替换任何K ，修改后的pos = pos –(K + 1)的值。由于K∈[1，步长] ，因此pos∈[pos –步长– 1，pos – 2] 。
  • 显然， pos – step 。如果N ，则选择相应的K = pos – N – 1并将K替换为-1并直接到达点N。
  • 如果N + 1 = pos ，则只需要执行一次-1运算。

下面是上述方法的实现：

C++
// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the minimum steps
// required to reach N by either moving
// i steps forward or 1 steps backward
int minimumsteps(int N)
{
 
    // Stores the required count
    int steps = 0;
 
    // IF total moves required
    // is less than double of N
    while (steps * (steps + 1) < 2 * N) {
 
        // Update steps
        steps++;
    }
 
    // Steps required to reach N
    if (steps * (steps + 1) / 2 == N + 1) {
 
        // Update steps
        steps++;
    }
 
    cout << steps;
}
 
// Driver Code
int main()
{
 
    // Given value of N
    int N = 18;
    minimumsteps(N);
}

Java
// Java program for the above approach
import java.util.*;
class GFG{
       
// Function to find the minimum steps
// required to reach N by either moving
// i steps forward or 1 steps backward
static void minimumsteps(int N)
{
 
    // Stores the required count
    int steps = 0;
 
    // IF total moves required
    // is less than double of N
    while (steps * (steps + 1) < 2 * N)
    {
 
        // Update steps
        steps++;
    }
 
    // Steps required to reach N
    if (steps * (steps + 1) / 2 == N + 1)
    {
 
        // Update steps
        steps++;
    }
 
    System.out.println(steps);
}
   
// Driver code
public static void main(String[] args)
{
   
    // Given value of N
    int N = 18;
    minimumsteps(N);
}
}
 
// This code is contributed by code_hunt.

Python3
# Function to find the minimum steps
# required to reach N by either moving
# i steps forward or 1 steps backward
 
def minimumsteps(N) :
 
    # Stores the required count
    steps = 0
 
    # IF total moves required
    # is less than double of N
    while (steps * (steps + 1) < 2 * N) :
 
        # Update steps
        steps += 1
 
    # Steps required to reach N
    if (steps * (steps + 1) / 2 == N + 1) :
 
        # Update steps
        steps += 1
    print(steps)
 
 
# Driver code
N = 18;
minimumsteps(N)
 
# This code is contributed by Dharanendra L V

C#
// C# program for the above approach
using System;
 
class GFG {
 
    // Function to find the minimum steps
    // required to reach N by either moving
    // i steps forward or 1 steps backward
    static void minimumsteps(int N)
    {
 
        // Stores the required count
        int steps = 0;
 
        // IF total moves required
        // is less than double of N
        while (steps * (steps + 1) < 2 * N) {
 
            // Update steps
            steps++;
        }
 
        // Steps required to reach N
        if (steps * (steps + 1) / 2 == N + 1) {
 
            // Update steps
            steps++;
        }
 
        Console.WriteLine(steps);
    }
 
    // Driver code
    static public void Main()
    {
 
        // Given value of N
        int N = 18;
        minimumsteps(N);
    }
}
 
// This code is contributed by Dharanendra LV

Javascript

输出： 
6

时间完成度： O(sqrt(N))
辅助空间： O(1)