LeetCode908 Smallest Range I(最小差值 I)JAVA實現
阿新 • • 發佈:2018-12-20
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
給定一個整數陣列 A
,對於每個整數 A[i]
,我們可以選擇任意 x
滿足 -K <= x <= K
x
加到 A[i]
中。
在此過程之後,我們得到一些陣列 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 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:0 解釋:B = [3,3,3] 或 B = [4,4,4]
提示:
1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000
解題思路:找出陣列A中的最大值max和最小值min,如果max和min的差值小於2K,那麼說明max和min可以化為相同的值,返回0.否則max-2和min+2,即返回max-min-2K。
JAVA實現程式碼:
class Solution {
public int smallestRangeI(int[] A, int K) {
if(A.length==1)//A的陣列大小為1直接返回
return 0;
int min=A[0],max=A[0];//記錄最大值 最小值
for(int i=1;i<A.length;i++)
{
if(A[i]>max)
max=A[i];
else if(A[i]<min)
min=A[i];
}
if((max-min)<=2*K)
return 0;
return max-min-2*K;
}
}