效率明显提高：先对整数进行排序，然后再查看相邻的整数。任何数字都会向上或向下最接近其邻居。

One obvious efficiency improvement: sort the integers first, then you can look at adjacent ones. Any number is going to be closest to its neighbour either up or down.

这会改变O的复杂度（n ² ）到O（n log n）。对于 n 的较小值，它不会产生明显的变化，但是就理论上的复杂性而言，这很重要。

That changes the complexity from O(n²) to O(n log n). Admittedly for the small value of n shown it's not going to make a significant difference, but in terms of theoretical complexity it's important.

您可能需要进行一次微优化：使用局部变量存储 Math的结果。 abs ，那么如果结果小于最小值，则无需重新计算。您可能想使用 dMin = Math.min（dMin，Math.abs（a [i]-a [j]））。

One micro-optimization you may want to make: use a local variable to store the result of Math.abs, then you won't need to recompute it if that turns out to be less than the minimum. Alternatively, you might want to use dMin = Math.min(dMin, Math.abs(a[i] - a[j])).

请注意，您需要注意边界条件-如果您允许使用负数，则减法可能会溢出。

Note that you need to be careful of border conditions - if you're permitting negative numbers, your subtraction might overflow.