LeetCode之Smallest Range I（Kotlin）-优快云博客

博客围绕一个算法问题展开，给定整数数组A，对每个元素A[i]可加一个范围在 -K 到 K 之间的 x 得到数组B，要找出B中最大值与最小值的最小差值。方法是先取数组最大、最小值，若二者差值大于2倍K，最小差为max - min - 2*K，否则为0。

问题： Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i]. After this process, we have some array B. Return the smallest possible difference between the maximum value of B and the minimum value of B.

Example 1:

Input: A = [1], K = 0
Output: 0
Explanation: B = [1]
Example 2:

Input: A = [0,10], K = 2
Output: 6
Explanation: B = [2,8]
Example 3:

Input: A = [1,3,6], K = 3
Output: 0
Explanation: B = [3,3,3] or B = [4,4,4]
 

Note:

1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000
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方法：先取出数组的最大值，然后取出数组的最小值，如果最大值和最小值的差比K的2倍还要大则最小差为max-min-2*K，否则为0，因为最小差即为0。

具体实现：

class SmallestRangeI {
    fun smallestRangeI(A: IntArray, K: Int): Int {
        val max = A.max()
        val min = A.min()
        if (max == null || min == null) {
            return 0
        }
        val result = if ((max - min - K * 2) >= 0) (max - min - K * 2) else 0
        return result
    }
}

fun main(args: Array<String>) {
    val A = intArrayOf(0, 10)
    val K = 2
    val smallestRangeI = SmallestRangeI()
    println(smallestRangeI.smallestRangeI(A, K))
}
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有问题随时沟通

具体代码实现可以参考Github