[英]How to calculate number of comparisons in merge sort?
這是我的算法。 我曾嘗試將numCompares
和numMoves
放置在許多不同的位置,但無法在可以看到有效結果的地方找到它。 比較= 2個整數正在彼此移動求值=一個元素已從其位置移出。 因此,交換將是2步。
/**
* Internal method that merges two sorted halves of a subarray.
* @param a an array of Comparable items.
* @param tmpArray an array to place the merged result.
* @param leftPos the left-most index of the subarray.
* @param rightPos the index of the start of the second half.
* @param rightEnd the right-most index of the subarray.
*/
private static void merge( int[] a, int[ ] tmpArray, int leftPos, int rightPos, int rightEnd )
{
int leftEnd = rightPos - 1; //left's last position
int tmpPos = leftPos; //left-most
int numElements = rightEnd - leftPos + 1;
// Main loop
while( leftPos <= leftEnd && rightPos <= rightEnd ){ //left and right pointers don't meet limit
//numCompares = numCompares+1; //+2 because of 2 conditions
if(a[leftPos] <= a[rightPos]){ //if left half element <= right half element
tmpArray[ tmpPos++ ] = a[ leftPos++ ]; //copy left side elements
}else{
tmpArray[ tmpPos++ ] = a[ rightPos++ ]; //copy right side elements
}
numMoves++;
numCompares = numCompares+2;
}
while( leftPos <= leftEnd ){ // Copy rest of first half while left is <= left end
tmpArray[ tmpPos++ ] = a[ leftPos++ ];
numCompares++;
numMoves++;
}
while( rightPos <= rightEnd ){ // Copy rest of right half while right is < right-most
tmpArray[ tmpPos++ ] = a[ rightPos++ ];
numCompares++;
numMoves++;
}
// Copy tmpArray back
for( int i = 0; i < numElements; i++, rightEnd-- ){
a[ rightEnd ] = tmpArray[ rightEnd ];
}
}
鑒於您想要最大的精度,這將是我的方法:
/**
* Internal method that merges two sorted halves of a subarray.
* @param a an array of Comparable items.
* @param tmpArray an array to place the merged result.
* @param leftPos the left-most index of the subarray.
* @param rightPos the index of the start of the second half.
* @param rightEnd the right-most index of the subarray.
*/
private static void merge( int[] a, int[ ] tmpArray, int leftPos, int rightPos, int rightEnd )
{
int leftEnd = rightPos - 1;
int tmpPos = leftPos;
int numElements = rightEnd - leftPos + 1;
while( leftPos <= leftEnd){
numCompares++; // this is for leftPos <= leftEnd
if(rightPos <= rightEnd) {
numCompares++; // this is for rightPos <= rightEnd
if (a[leftPos] <= a[rightPos]) { //if left half element <= right half element
tmpArray[tmpPos++] = a[leftPos++]; //copy left side elements
} else {
tmpArray[tmpPos++] = a[rightPos++]; //copy right side elements
}
numMoves++;
numCompares++; // this is for a[leftPos] <= a[rightPos]
} else {
break;
}
}
numCompares++; //The while loop exited. This has happened because (leftPos <= leftEnd) or (rightPos <= rightEnd) failed.
while( leftPos <= leftEnd ){ // Copy rest of first half while left is <= left end
tmpArray[ tmpPos++ ] = a[ leftPos++ ];
numCompares++;
numMoves++;
}
numCompares++; //The while loop exited. This has happened because leftPos <= leftEnd failed.
while( rightPos <= rightEnd ){ // Copy rest of right half while right is < right-most
tmpArray[ tmpPos++ ] = a[ rightPos++ ];
numCompares++;
numMoves++;
}
numCompares++; //The while loop exited. This has happened because rightPos <= rightEnd failed.
// Copy tmpArray back
// I assume that you don't want account for this operation in your measurements, since you didn't try to do it yourself?
for( int i = 0; i < numElements; i++, rightEnd-- ){
a[ rightEnd ] = tmpArray[ rightEnd ];
// numCompares++; // This is for (i < numElements)
// numMoves++;
}
// numCompares++; //The for loop exited. This has happened because (i < numElements) failed.
}
如果將其與漸近運行時間進行比較,請嘗試逐漸增加大小並繪制測量圖。 它不太適合小樣本量。 另外,此代碼還計算了數組中的寫入量。 如果您也要考慮讀取,則需要在numMoves遞增的地方添加兩個而不是一個。
希望能幫助到你。 祝好運 :)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.