[英]MergeSort and QuickSort swaps questions
In quickSort, given an array a[] = {1, 2, 3, 4, 5};
在快速排序中,给定一个数组
a[] = {1, 2, 3, 4, 5};
when I count swaps, it always returns 5
when it's sorted.当我计算交换时,它在排序时总是返回
5
。 How is that?那个怎么样? I thought that swaps should be counted only when it actually swaps two numbers.
我认为只有在实际交换两个数字时才应计算交换。 Another example - given an array
a[] = {2, 1, 3, 4, 5};
另一个例子 - 给定一个数组
a[] = {2, 1, 3, 4, 5};
it returns swaps = 4
.它返回
swaps = 4
。 I don't really understand how they work.我真的不明白他们是如何工作的。
int swapsQuick = 0;
int partition(int *L, int left, int right) {
int pivot = left;
int p_val = L[pivot];
while (left < right) {
while (L[left] <= p_val)
left++;
while (L[right] > p_val)
right--;
if (left < right) {
swap(&L[left], &L[right]);
swapsQuick++;
}
}
swap(&L[pivot], &L[right]);
swapsQuick++;
return right;
}
int quicksort(int *L, int start, int end) {
if (start >= end)
return swapsQuick;
int splitPoint = partition(L, start, end);
quicksort(L, start, splitPoint - 1);
quicksort(L, splitPoint + 1, end);
return swapsQuick;
}
Also, which part of this mergeSort
help function actually swaps two numbers?另外,这个
mergeSort
的哪一部分帮助 function 实际上交换了两个数字?
void merge(int arr[], int l, int m, int r) {
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
int L[n1], R[n2];
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
i = 0;
j = 0;
k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
} else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}
There is a problem in your partition
function that may explain the incorrect count:您的
partition
function 存在问题,这可能解释了不正确的计数:
the loop while (L[left] <= p_val) left++;
循环
while (L[left] <= p_val) left++;
may go beyond the end of the array if the pivot chosen is the smallest element of the slice, which is always the case for a sorted slice.如果选择的 pivot 是切片的最小元素,则 go 可能会超出数组的末尾,对于排序切片来说总是如此。 You should change it to:
您应该将其更改为:
while (left <= right && L[left] <= p_val) left++;
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