我有一个递归算法来计算加权中位数。我试图弄清楚Big-O的时间复杂度是什么,但我有点被卡住了。有谁可以帮助我吗。谢谢大家的帮助。这是JAVA中的代码:
public static double WeightedMedian(ArrayList<Double> a1, ArrayList<Double> a2, int p, int r) {
if (r == p) {
return a1.get(p);
}
if (r-p == 0) {
if (a2.get(p) == a2.get(r)) {
return (a1.get(p) + a1.get(r))/2;
}
if (a2.get(p) > a2.get(r)) {
return a1.get(p);
} else {
return a1.get(r);}
}
long q = partition(a1, p, r);
double wl=0,wg=0;
for (int i=p; i<=q-1; i++) {
wl += a2.get(i);
}
for (int i=(int) (q+1); i<=r; i++) {
wg += a2.get(i);
}
if (wl<0.5 && wg<0.5) {
return a1.get((int)q);
} else {
if (wl > wg) {
double x = a2.get((int)q) + wg;
a2.set((int) q,x);
return WeightedMedian(a1,a2, p+1, (int)q);
} else {
double x = a2.get((int)q) + wl;
a2.set((int) q,x);
return WeightedMedian(a1, a2, (int)q, r);
}
}
对不起,这是我第一次在这里发表文章,所以我不太喜欢尝试格式化更好的代码,但是它总是在奇怪的地方等等。总之,分区代码如下:
public static long partition (ArrayList<Double> arr, int low, int high)
{
double pivot = arr.get(high);
int i = low - 1;
for (int j = low; j <= high- 1; j++)
{
if (arr.get(j) <= pivot)
{
i++;
double temp = arr.get(i);
arr.set(i,arr.get(j));
arr.set(j,temp);
}
}
double temp1 = arr.get(i + 1);
arr.set(i + 1, arr.get(high));
arr.set(high,temp1);
return i + 1;
}
答案 0 :(得分:1)
public static double WeightedMedian(ArrayList<Double> a1, ArrayList<Double> a2, int p, int r) {
if (r == p) {
return a1.get(p); //O(1)
}
if (r-p == 0) {
if (a2.get(p) == a2.get(r)) { //O(1) + O(1)
return (a1.get(p) + a1.get(r))/2; //O(1) + O(1)
}
if (a2.get(p) > a2.get(r)) { //O(1) + O(1)
return a1.get(p); //O(1)
} else {
return a1.get(r);} //O(1)
}
long q = partition(a1, p, r);
double wl=0,wg=0;
for (int i=p; i<=q-1; i++) { //O(n)
wl += a2.get(i);
}
for (int i=(int) (q+1); i<=r; i++) { //O(n)
wg += a2.get(i);
}
if (wl<0.5 && wg<0.5) {
return a1.get((int)q); //O(1)
} else {
if (wl > wg) {
double x = a2.get((int)q) + wg; //O(1)
a2.set((int) q,x); //O(1)
return WeightedMedian(a1,a2, p+1, (int)q);
} else {
double x = a2.get((int)q) + wl; //O(1)
a2.set((int) q,x); //O(1)
return WeightedMedian(a1, a2, (int)q, r);
}
}
所以上面的方法是O(n),从O(1)... + O(n)= O(n)
public static long partition (ArrayList<Double> arr, int low, int high) {
double pivot = arr.get(high); //O(1)
int i = low - 1;
for (int j = low; j <= high- 1; j++) //O(n)
{
if (arr.get(j) <= pivot) //O(1)
{
i++;
double temp = arr.get(i); //O(1)
arr.set(i,arr.get(j)); //O(1)
arr.set(j,temp); //O(1)
}
}
double temp1 = arr.get(i + 1); //O(1)
arr.set(i + 1, arr.get(high)); //O(1)
arr.set(high,temp1); //O(1)
return i + 1;
}
上述方法“分区”,也是O(n),由O(1)... + O(n)导出
所以O(n),因为arraylist是使用索引的直接访问,并且所有的get / sets都是O(1),并且所有循环都是O(n)