o1和o2分别表示在一个数组中的前一个数和后一个数。
如果像按照升序的条件排列,就要return o1-o2; (前面的数 - 后面的数);
解释如下:
降序也是同样的道理。
JDK中关于这部分的源码:
public static <T> void sort(T[] a, Comparator<? super T> c) { if (c == null) { sort(a); } else { if (LegacyMergeSort.userRequested) legacyMergeSort(a, c); else TimSort.sort(a, 0, a.length, c, null, 0, 0); } }
进入 legacyMergeSort;
private static <T> void legacyMergeSort(T[] a, Comparator<? super T> c) { T[] aux = a.clone(); if (c==null) mergeSort(aux, a, 0, a.length, 0); else mergeSort(aux, a, 0, a.length, 0, c); }
进入mergeSort();
private static void mergeSort(Object[] src, Object[] dest, int low, int high, int off, Comparator c) { int length = high - low; // Insertion sort on smallest arrays if (length < INSERTIONSORT_THRESHOLD) { for (int i=low; i<high; i++) for (int j=i; j>low && c.compare(dest[j-1], dest[j])>0; j--) swap(dest, j, j-1); return; } // Recursively sort halves of dest into src int destLow = low; int destHigh = high; low += off; high += off; int mid = (low + high) >>> 1; mergeSort(dest, src, low, mid, -off, c); mergeSort(dest, src, mid, high, -off, c); // If list is already sorted, just copy from src to dest. This is an // optimization that results in faster sorts for nearly ordered lists. if (c.compare(src[mid-1], src[mid]) <= 0) { System.arraycopy(src, low, dest, destLow, length); return; } // Merge sorted halves (now in src) into dest for(int i = destLow, p = low, q = mid; i < destHigh; i++) { if (q >= high || p < mid && c.compare(src[p], src[q]) <= 0) dest[i] = src[p++]; else dest[i] = src[q++]; } }
其中关于交换两个元素的位置:
if (length < INSERTIONSORT_THRESHOLD) { for (int i=low; i<high; i++) for (int j=i; j>low && c.compare(dest[j-1], dest[j])>0; j--) swap(dest, j, j-1); return; }