910 Smallest Range II 最小差值 II
Description:
You are given an integer array nums and an integer k.
For each index i where 0 <= i < nums.length, change nums[i] to be either nums[i] + k or nums[i] - k.
The score of nums is the difference between the maximum and minimum elements in nums.
Return the minimum score of nums after changing the values at each index.
Example:
Example 1:
Input: nums = [1], k = 0
Output: 0
Explanation: The score is max(nums) - min(nums) = 1 - 1 = 0.
Example 2:
Input: nums = [0,10], k = 2
Output: 6
Explanation: Change nums to be [2, 8]. The score is max(nums) - min(nums) = 8 - 2 = 6.
Example 3:
Input: nums = [1,3,6], k = 3
Output: 3
Explanation: Change nums to be [4, 6, 3]. The score is max(nums) - min(nums) = 6 - 3 = 3.
Constraints:
1 <= nums.length <= 10^4
0 <= nums[i] <= 10^4
0 <= k <= 10^4
題目描述:
給你一個整數數組 A，對於每個整數 A[i]，可以選擇 x = -K 或是 x = K （K 總是非負整數），並將 x 加到 A[i] 中。
在此過程之後，得到數組 B。
返回 B 的最大值和 B 的最小值之間可能存在的最小差值。
示例 :
示例 1：
輸入：A = [1], K = 0
輸出：0
解釋：B = [1]
示例 2：
輸入：A = [0,10], K = 2
輸出：6
解釋：B = [2,8]
示例 3：
輸入：A = [1,3,6], K = 3
輸出：3
解釋：B = [4,6,3]
提示:
1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000
思路:
排序
將 nums 分成兩個部分, 前一半需要加上 k, 後一半需要減去 k
取上述兩個部分的最小差值即可
時間複雜度爲 O(nlgn), 空間複雜度爲 O(1)
代碼:
C++:
class Solution
{
public:
int smallestRangeII(vector<int>& nums, int k)
{
sort(nums.begin(), nums.end());
int n = nums.size(), result = nums.back() - nums.front();
for (int i = 1; i < n; i++) result = min(result, max(nums[i - 1] + k, nums.back() - k) - min(nums.front() + k, nums[i] - k));
return result;
}
};
Java:
class Solution {
public int smallestRangeII(int[] nums, int k) {
Arrays.sort(nums);
int n = nums.length, result = nums[n - 1] - nums[0];
for (int i = 1; i < n; i++) result = Math.min(result, Math.max(nums[i - 1] + k, nums[n - 1] - k) - Math.min(nums[0] + k, nums[i] - k));
return result;
}
}
Python:
class Solution:
def smallestRangeII(self, nums: List[int], k: int) -> int:
nums.sort()
n, result = len(nums), nums[-1] - nums[0]
for i in range(1, n):
result = min(result, max(nums[i - 1] + k, nums[-1] - k) - min(nums[0] + k, nums[i] - k))
return result