0910. Smallest Range II (M)

Smallest Range II (M)

題目

Given an array A of integers, for each integer A[i] we need to choose either x = -K or x = K, and add x to A[i] (only once).

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: 3
Explanation: B = [4,6,3]

Note:

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

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題意

遍歷一個數組，對每一個值進行加K或減K的操做，求獲得的新數組中(max-min)的最小值。

思路

由於每一個元素都必須加K或者減K，能夠將數組分紅兩組，一組全加K，一組全減K，很明顯要讓小的數加K，讓大的數減K。所以能夠先對數組排序，而後肯定一個分界點，讓左邊的子數組left所有加K，右邊的子數組right所有減K；同時注意到，兩個子數組中的最小值都是各自的第一個元素，最大值都是各自的最後一個元素，所以能夠比較得出全局的最大值和最小值，再以此更新結果。

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代碼實現

Java

class Solution {
    public int smallestRangeII(int[] A, int K) {
        int ans = Integer.MAX_VALUE;
        Arrays.sort(A);

        for (int len = 0; len <= A.length; len++) {
            int max = 0, min = 0;
            if (len == 0 || len == A.length) {
                max = A[A.length - 1];
                min = A[0];
            } else  {
                max = Math.max(A[len - 1] + K, A[A.length - 1] - K);
                min = Math.min(A[0] + K, A[len] - K);
            }
            ans = Math.min(ans, max - min);
        }

        return ans;
    }
}