[英]Big O complexity of recursive algorithm
I have a recursive algorithm for the calculation of the weighted median.我有一个递归算法来计算加权中位数。 I'm trying to figure out what the Big-O time complexity will be but i am kinda stuck.我试图弄清楚 Big-O 时间复杂度是多少,但我有点卡住了。 Can someone help me please.有人能帮助我吗。 Thank you all for the help.谢谢大家的帮助。 Here is the code in JAVA: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);
}
}
sorry this is my first time posting here so im not really thta good i tried to format the code better but it kept going in strange places etc. Anyway partition code is as follows:抱歉,这是我第一次在这里发帖,所以我不是很好,我试图更好地格式化代码,但它一直在奇怪的地方等。无论如何分区代码如下:
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;
}
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);
}
}
So the above method is O(n), from O(1)...+O(n) = O(n)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;
}
And the above method "partition", also O(n), derived by O(1)... + O(n)而上面的方法“分区”,也是O(n),由O(1)... + O(n)导出
So, O(n), as arraylist is direct access with index's and all get/sets are O(1) and all of your loops are O(n)所以,O(n),因为 arraylist 是直接访问索引的,所有获取/集都是 O(1),所有循环都是 O(n)
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