插入排序,比较次数
Insertion sort, number of comparisons
大家好,对于一项作业,我们必须计算多种算法的比较次数。我使用的是 Sedgewick & Wayne 所著 "Algorithms" 一书中的代码。我实际上没有看到我的代码哪里错了......一旦我们要比较一些东西我就会计算我的比较......
public long sort(Comparable[] a) {
if (a == null) {
throw new IllegalArgumentException("argument 'array' must not be null.");
}
int N = a.length;
for (int i = 0; i < N; i++) {
for (int j = i; j > 0; j--) {
this.comparisons++;
if(less(a[j], a[j-1]))
exch(a, j, j-1);
}
assert isSorted(a, 0, i);
}
assert isSorted(a);
return this.comparisons;
}
我用的less方法:
private boolean less(Comparable v, Comparable w) {
return (v.compareTo(w) < 0);
}
必须通过这个测试
Integer[] array = {4, 2, 1, 3, -1};
Comparable[] arrayClone1 = array.clone();
Comparable[] arrayClone2 = array.clone();
long nbCompares1 = i.sort(arrayClone1);
long nbCompares2 = i.sort(arrayClone2);
System.out.println("1" + nbCompares1);
System.out.println("2" + nbCompares2);
这两个应该相等....
isSorted 方法:
private boolean isSorted(Comparable[] a) {
System.out.println("here");
return isSorted(a, 0, a.length - 1);
}
// is the array sorted from a[lo] to a[hi]
private boolean isSorted(Comparable[] a, int lo, int hi) {
System.out.println("here1");
for (int i = lo + 1; i <= hi; i++)
if (less(a[i], a[i-1])) return false;
return true;
}
有人对此有想法吗?帮助将不胜感激!
比较次数应该正好是N*(N-1)/2。也许你在其他地方弄乱了 comparisons
字段,所以我建议改用局部变量:
public long sort(Comparable[] a) {
if (a == null) {
throw new IllegalArgumentException("argument 'array' must not be null.");
}
int N = a.length;
int comparisonsCount = 0; // use this instead
for (int i = 0; i < N; i++) {
for (int j = i; j > 0; j--) {
comparisonsCount++; // edit here
if(less(a[j], a[j-1]))
exch(a, j, j-1);
}
assert isSorted(a, 0, i);
}
assert isSorted(a);
return comparisonsCount; // and here
}
大家好,对于一项作业,我们必须计算多种算法的比较次数。我使用的是 Sedgewick & Wayne 所著 "Algorithms" 一书中的代码。我实际上没有看到我的代码哪里错了......一旦我们要比较一些东西我就会计算我的比较......
public long sort(Comparable[] a) {
if (a == null) {
throw new IllegalArgumentException("argument 'array' must not be null.");
}
int N = a.length;
for (int i = 0; i < N; i++) {
for (int j = i; j > 0; j--) {
this.comparisons++;
if(less(a[j], a[j-1]))
exch(a, j, j-1);
}
assert isSorted(a, 0, i);
}
assert isSorted(a);
return this.comparisons;
}
我用的less方法:
private boolean less(Comparable v, Comparable w) {
return (v.compareTo(w) < 0);
}
必须通过这个测试
Integer[] array = {4, 2, 1, 3, -1};
Comparable[] arrayClone1 = array.clone();
Comparable[] arrayClone2 = array.clone();
long nbCompares1 = i.sort(arrayClone1);
long nbCompares2 = i.sort(arrayClone2);
System.out.println("1" + nbCompares1);
System.out.println("2" + nbCompares2);
这两个应该相等....
isSorted 方法:
private boolean isSorted(Comparable[] a) {
System.out.println("here");
return isSorted(a, 0, a.length - 1);
}
// is the array sorted from a[lo] to a[hi]
private boolean isSorted(Comparable[] a, int lo, int hi) {
System.out.println("here1");
for (int i = lo + 1; i <= hi; i++)
if (less(a[i], a[i-1])) return false;
return true;
}
有人对此有想法吗?帮助将不胜感激!
比较次数应该正好是N*(N-1)/2。也许你在其他地方弄乱了 comparisons
字段,所以我建议改用局部变量:
public long sort(Comparable[] a) {
if (a == null) {
throw new IllegalArgumentException("argument 'array' must not be null.");
}
int N = a.length;
int comparisonsCount = 0; // use this instead
for (int i = 0; i < N; i++) {
for (int j = i; j > 0; j--) {
comparisonsCount++; // edit here
if(less(a[j], a[j-1]))
exch(a, j, j-1);
}
assert isSorted(a, 0, i);
}
assert isSorted(a);
return comparisonsCount; // and here
}