给定无限线上的目标位置(从-无限到+无限)。从0表格开始，您必须按照以下说明进行移动才能到达目标：在第ii步中，您可以向前或向后执行i步。找到达到目标所需的最小移动量。
例子 ：
Input : target = 3
Output : 2
Explanation:
On the first move we step from 0 to 1.
On the second step we step from 1 to 3.
Input: target = 2
Output: 3
Explanation:
On the first move we step from 0 to 1.
On the second move we step from 1 to -1.
On the third move we step from -1 to 2.方法 ：
这个想法类似于此处的O(n)方法中讨论的。
继续添加总和= 1 + 2 + .. + n> =目标。求解该二次方程式可得出最小的n，使得总和> =目标，即求解n(n + 1)/ 2中的n –目标> = 0可获得最小的n。
如果sum ==目标，则答案为n。现在是下一个总和大于目标的情况。通过索引比目标(即和目标)领先多少步来找到差异。
情况1：差值等于n的答案是(因为总是会发生移动翻转，这将导致目标移动)。
情况2：差是奇数，然后再走一步，即将n + 1加到和，然后再次取差。如果差异是n + 1，则答案又是另一步，这肯定会造成差异，即使答案是n + 2。
说明：由于差异是奇数。目标是奇数或偶数。
情况1：n为偶数(1 + 2 + 3 +…+ n)，然后加n + 1会使差为偶数。
情况2：n为奇数，然后加n + 1不会有什么区别，即使如此，又再走一步，即n + 2。
C++
// CPP code to find minimum moves
// to reach target
#include
using namespace std;
// Function to find minimum steps
// to reach target
int StepstoReachTarget(int target)
{
// Handling negatives
// by symmetry
target = abs(target);
// Keep moving while sum is
// smaller i.e calculating n
int n = ceil((-1.0 + sqrt(1 + 8.0 * target)) / 2);
int sum = n * (n + 1) / 2;
if (sum == target)
return n;
int d = sum - target;
// case 1 : d is even
if ((d & 1) == 0)
return n;
// d is odd
else
return n + ((n & 1) ? 2 : 1);
}
// Driver code
int main()
{
int target = 5;
// Function call
cout << StepstoReachTarget(target);
return 0;
} Java
// Java code to find minimum moves
// to reach target
import java.lang.*;
class GFG {
// Function to find minimum steps
// to reach target
static int StepstoReachTarget(int target)
{
// Handling negatives
// by symmetry
target = Math.abs(target);
// Keep moving while sum is
// smaller i.e calculating n
int n = (int)Math.ceil(
(-1.0 + (int)Math.sqrt(1 + 8.0 * target)) / 2);
int sum = n * (n + 1) / 2;
if (sum == target)
return n;
int d = sum - target;
// case 1 : d is even
if ((d & 1) == 0)
return n;
// d is odd
else
return n + ((n & 1) != 0 ? 2 : 1);
}
// Driver code
public static void main(String[] arg)
{
int target = 5;
// Function call
System.out.println(StepstoReachTarget(target));
}
}
// This code is contributed by
// Smitha Dinesh SemwalPython3
# Python code to find minimum
# moves to reach target
import math
# Function to find minimum
# steps to reach target
def StepstoReachTarget(target):
# Handling negatives
# by symmetry
target = abs(target)
# Keep moving while sum is
# smaller i.e calculating n
n = math.ceil((-1.0 + math.sqrt(1 +
8.0 * target)) / 2)
sum = n * (n + 1) / 2
if (sum == target):
return n
d = sum - target
# case 1 : d is even
if ((int(d) & 1) == 0):
return n
# d is odd
else:
if(int(d) & 1):
return n + 2
return n + 1
# Driver code
target = 5
# Function call
print(StepstoReachTarget(target))
# This code is contributed by
# Manish Shaw(manishshaw1)C#
// C# code to find minimum moves
// to reach target
using System;
class GFG {
// Function to find minimum steps
// to reach target
static int StepstoReachTarget(int target)
{
// Handling negatives
// by symmetry
target = Math.Abs(target);
// Keep moving while sum is
// smaller i.e calculating n
int n = (int)Math.Ceiling(
(-1.0 + (int)Math.Sqrt(1 + 8.0 * target)) / 2);
int sum = n * (n + 1) / 2;
if (sum == target)
return n;
int d = sum - target;
// case 1 : d is even
if ((d & 1) == 0)
return n;
// d is odd
else
return n + ((n & 1) != 0 ? 2 : 1);
}
// Driver code
public static void Main()
{
int target = 5;
// Function call
Console.Write(StepstoReachTarget(target));
}
}
// This code is contributed by nitin mittal.PHP
Javascript
输出 ：
5