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  • 908. Smallest Range I

    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. 1 <= A.length <= 10000
    2. 0 <= A[i] <= 10000
    3. 0 <= K <= 10000

    Approach #1: Math. [Java]

    class Solution {
        public int smallestRangeI(int[] A, int K) {
            int min = Integer.MAX_VALUE;
            int max = Integer.MIN_VALUE;
            for (int i = 0; i < A.length; ++i) {
                if (min > A[i]) min = A[i];
                if (max < A[i]) max = A[i];
            }
            if (max - min < 2 * K) return 0;
            return max - min - 2 * K;
        }
    }
    

      

    Analysis:

    If you can find that the result only relate with (max - min) and 2 * K, it will become easy to solve.

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/10914741.html
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