[英]Count Inversions with merge sort in C++
我正在開發我的C ++技巧的前幾種算法,目前正在編碼一種使用歸並排序計算反轉的方法。 我設法使工作合並排序在一起,但是在跟蹤反轉次數時遇到了一些麻煩。 有什么想法從這里去嗎? 我該如何跟蹤像這樣的遞歸算法的反轉次數? 另外,我在互聯網旅行中看到了幾種不同的實現方式,發現大多數人都偏離了std :: vector方法,知道為什么嗎? 感謝您的幫助,下面是我的代碼!
#include <iostream>
#include <math.h>
#include <vector>
using namespace std;
vector<int> print(vector<int> input){
for(int i=0; i<input.size(); i++){
cout<<input[i]<<",";
}
cout<<endl;
return input;
}
vector<int> merge(vector<int> left,vector<int> right){
//set up some varibles
vector<int> output;
int i=0;
int j=0;
//loop through the lists and merge
while(i<left.size() && j<right.size()){
//push the smallest of the two to the vector output
if(left[i]<=right[j]){
output.push_back(left[i]);
i+=1;
}
if(left[i]>right[i]){
output.push_back(right[j]);
j+=1;
}
}
//push the remnants of the vectors to output
for(i; i<left.size(); i++){
output.push_back(left[i]);
}
for(j; j<right.size(); j++){
output.push_back(right[j]);
}
return output;
}//end merge
vector<int> merge_sort(vector<int> input){
//check the size of the vector
if(input.size()<2){
return input;
}
else{
//int new vectors
vector<int> left;
vector<int> right;
vector<int> output;
//find the middle of the input vector
int middle=(input.size())/2;
//build the left vector
for(int i=0; i<middle; i++){
left.push_back(input[i]);
}
//build the right vector
for(int i=middle; i<input.size(); i++){
right.push_back(input[i]);
}
//make recursive calls
left=merge_sort(left);
right=merge_sort(right);
//call merge
output=merge(left,right);
return output;
}
}
int main()
{
vector<int> output;
vector<int> input;
input.push_back(2);
input.push_back(1);
input.push_back(10);
input.push_back(4);
output=merge_sort(input);
print(output);
}
好消息:從這里開始計算反轉非常容易。
考慮一下您的“合並”方法。 每次將左側向量中的元素放入輸出中時,都不會更改其相對於右側元素的位置。 另一方面,每當您從右向量中添加一個元素時,您就將其放置在左向量中之前仍要處理的所有元素的“之前”,即先將它們放置在“之后”,即創建(left.size- i)“倒置”。
如果需要,您可以通過歸納輕松證明這一點。
因此答案很簡單:將int *傳遞給您的merge方法,並在每次從右側向量中推送元素時將其遞增(left.size-i)。
編輯:工作代碼示例
#include <iostream>
#include <vector>
// removed useless dependency math.h
using namespace std;
// void type -> does not return anything
void print (vector<int> input) {
// range-based for loop (since C++ 11)
// no brackets -> only one instruction in for loop
for(int i : input)
cout << i << ",";
}
vector<int> merge (vector<int> left, vector<int> right, int * inv_count) {
vector<int> output;
// multiple variable definition of the same type
int i=0, j=0;
// spaces around "<", after "while", before "{" for readability
while (i < left.size() && j < right.size()) {
// one-instruction trick again
if (left[i] <= right[j])
// i++ is evaluated to <previous value of i> and then increments i
// this is strictly equivalent to your code, but shorter
// check the difference with ++i
output.push_back(left[i++]);
// else because the two conditions were complementary
else {
output.push_back(right[j++]);
// pointer incrementation
*inv_count += (left.size() - i);
}
}
// first field of for ommited because there is no need to initialize i
for(; i < left.size(); i++)
output.push_back(left[i]);
for(; j < right.size(); j++)
output.push_back(right[j]);
return output;
}
vector<int> merge_sort (vector<int> input, int * inv_count) {
// no-braces-idiom again
// spaces around "<" and after "if" for readability
if (input.size() < 2)
return input;
// no need for else keyword because of the return
// multiple variable definition
vector<int> left, right;
int middle = input.size() / 2;
// one-instruction for loop
for(int i=0; i < middle; i++)
left.push_back(input[i]);
for(int i=middle; i < input.size(); i++)
right.push_back(input[i]);
// no need for intermediate variable
return merge( merge_sort(left, inv_count),
merge_sort(right, inv_count),
inv_count);
}
// consistent convention : brace on the same line as function name with a space
int main () {
// vector initialization (valid only since C++ 11)
vector<int> input = {2, 1, 10, 4, 42, 3, 21, 7};
int inv_count = 0;
// No need for intermediate variables again, you can chain functions
print( merge_sort(input, &inv_count) );
// The value inv_count was modified although not returned
cout << "-> " << inv_count << " inversions" << endl;
}
我修改了您的代碼,以包括一些常見的C ++習慣用法。 因為您使用的是C ++ 14標記,所以我還使用了自C ++ 11起才可用的技巧。 我不建議您在所有地方都使用所有這些技巧,因為它們是一種很好的學習體驗,所以此處不包含這些技巧。
我建議您在深入了解C ++之前先閱讀有關指針的知識。
還要注意,此代碼絕不是最佳選擇:創建了太多中間向量,並且向量在這里沒有用,數組就足夠了。 但是,我將再離開一次。
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