public class ComparableUser implements Comparable<Object> {
public String name;
public int age;
public ComparableUser(String name, int age) {
this.name = name;
this.age = age;
}
@Override
public int compareTo(Object o) {
return age - ((ComparableUser) o).age;
}
}
ComparableUser[] users = new ComparableUser[] {
new ComparableUser("Jim", 80),
new ComparableUser("Tom", 10),
new ComparableUser("Lily", 30),
new ComparableUser("Marry", 20)
};
Arrays.sort(users);[color=red]Comparable接口，只要类继承该接口，就可以实现对象的排序功能。[/color]
public class User {
public String name;
public int age;
public User(String name, int age) {
this.name = name;
this.age = age;
}
}
public class ComparatorUser implements Comparator<User> {
@Override
public int compare(User o1, User o2) {
return o1.age - o2.age;
}
}
User[] users2 = new User[] {
new User("Jim", 80),
new User("Tom", 10),
new User("Lily", 30),
new User("Marry", 20)
};
Arrays.sort(users2, new ComparatorUser());[color=red]Comparator接口，可以用来对已有的未继承Comparable接口的，进行排序。[/color]
sort方法，在JDK6和JDK7中实现方式不一样，具体如下：
在Java 6中Arrays.sort()和Collections.sort()使用的是MergeSort，而在Java 7中，内部实现换成了TimSort，其对对象间比较的实现要求更加严格：
Comparator的实现必须保证以下几点（出自这儿）：
a). sgn(compare(x, y)) == -sgn(compare(y, x))
b). (compare(x, y)>0) && (compare(y, z)>0) 意味着 compare(x, z)>0
c). compare(x, y)==0 意味着对于任意的z：sgn(compare(x, z))==sgn(compare(y, z)) 均成立
而我们的代码中，某个compare()实现片段是这样的：
public int compare(ComparatorTest o1, ComparatorTest o2) {
return o1.getValue() > o2.getValue() ? 1 : -1;
}这就违背了a)原则：假设X的value为1，Y的value也为1；那么compare(X, Y) ≠ –compare(Y, X)
PS: TimSort不仅内置在各种JDK 7的版本，也存在于Android SDK中（尽管其并没有使用JDK 7）。
下面看下sort方法中的具体排序源码
static void sort(Object[] a, int lo, int hi) {
rangeCheck(a.length, lo, hi);
int nRemaining = hi - lo;
if (nRemaining < 2)
return; // Arrays of size 0 and 1 are always sorted
// If array is small, do a "mini-TimSort" with no merges
if (nRemaining < MIN_MERGE) {
int initRunLen = countRunAndMakeAscending(a, lo, hi);
binarySort(a, lo, hi, lo + initRunLen);
return;
}
............................
}从代码中可以看出使用的是二分法排序binarySort
/**
* Sorts the specified portion of the specified array using a binary
* insertion sort. This is the best method for sorting small numbers
* of elements. It requires O(n log n) compares, but O(n^2) data
* movement (worst case).
*
* If the initial part of the specified range is already sorted,
* this method can take advantage of it: the method assumes that the
* elements from index {@code lo}, inclusive, to {@code start},
* exclusive are already sorted.
*
* @param a the array in which a range is to be sorted
* @param lo the index of the first element in the range to be sorted
* @param hi the index after the last element in the range to be sorted
* @param start the index of the first element in the range that is
* not already known to be sorted ({@code lo <= start <= hi})
*/
@SuppressWarnings("fallthrough")
private static void binarySort(Object[] a, int lo, int hi, int start) {
assert lo <= start && start <= hi;
if (start == lo)
start++;
for ( ; start < hi; start++) {
@SuppressWarnings("unchecked")
Comparable<Object> pivot = (Comparable) a[start];
// Set left (and right) to the index where a[start] (pivot) belongs
int left = lo;
int right = start;
assert left <= right;
/*
* Invariants:
* pivot >= all in [lo, left).
* pivot < all in [right, start).
*/
while (left < right) {
int mid = (left + right) >>> 1;
if (pivot.compareTo(a[mid]) < 0)
right = mid;
else
left = mid + 1;
}
assert left == right;
/*
* The invariants still hold: pivot >= all in [lo, left) and
* pivot < all in [left, start), so pivot belongs at left. Note
* that if there are elements equal to pivot, left points to the
* first slot after them -- that's why this sort is stable.
* Slide elements over to make room for pivot.
*/
int n = start - left; // The number of elements to move
// Switch is just an optimization for arraycopy in default case
switch (n) {
case 2: a[left + 2] = a[left + 1];
case 1: a[left + 1] = a[left];
break;
default: System.arraycopy(a, left, a, left + 1, n);
}
a[left] = pivot;
}
}
