2208. 将数组和减半的最少操作次数 力扣！7月26每日一题

**将数组和减半的最少操作次数**

**问题描述**

给定一个整数数组 `nums` 和一个目标值 `target`，请找到将 `nums` 变为 `target` 的最少操作次数。每个操作可以是将数字加一或减一。

**示例1**

输入：`nums = [4,3,2,1]`, `target =5`

输出：`1`

解释：将 `nums` 变为 `[4,3,2,2]`，然后再减一即可得到目标值 `5`。

**示例2**

输入：`nums = [1,2,3,4,5]`, `target =6`

输出：`0`

解释：`nums` 本身已经是目标值 `6`，所以不需要任何操作。

**示例3**

输入：`nums = [1,2,3,4,5]`, `target =7`

输出：`1`

解释：将 `nums` 变为 `[1,2,3,4,6]`，然后再减一即可得到目标值 `7`。

**解决方案**

我们可以使用贪婪算法来解决这个问题。贪婺算法的基本思想是，每一步都选择最优解，以期望达到全局最优解。

def minOperations(nums, target):
 """
 Returns the minimum number of operations to transform nums into target.

 :param nums: A list of integers.
 :type nums: List[int]
 :param target: The target value.
 :type target: int :return: The minimum number of operations.
 :rtype: int """
 # Calculate the total sum of nums total_sum = sum(nums)
 # If the total sum is equal to the target, return0 if total_sum == target:
 return0 # Initialize a list to store the minimum number of operations for each possible sum dp = [float('inf')] * (total_sum + target +1)
 # Base case: The minimum number of operations for sum0 is0 dp[0] =0 # Iterate over each possible sum from1 to total_sum + target for i in range(1, total_sum + target +1):
 # For each possible sum, iterate over each element in nums for num in nums:
 # If the current sum is greater than or equal to the current number, update the minimum number of operations if i >= num:
 dp[i] = min(dp[i], dp[i - num] +1)
 # Return the minimum number of operations for the target value return dp[target]