这是我的算法。我已尝试将numCompares
和numMoves
放在许多不同的位置,但我无法在可以看到有效结果的地方找到它。
比较=彼此评估的2个整数
move =一个元素已从其位置移开。因此交换将是2个动作。
/**
* 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 ];
}
}
答案 0 :(得分:1)
考虑到你想要最大精度,这将是我的方法:
/**
* 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.
}
如果要将其与渐近运行时间进行比较,请尝试逐渐增大尺寸并绘制测量值图表。它不太可能适合小样本量。 另外,作为旁注,此代码计算数组中的写入量。如果你也要考虑读取,你需要在每个地方添加两个而不是一个numMoves递增。
希望它有所帮助。祝你好运:)