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法學研究C/C++ 程式如何使用歸併排序計算陣列中的逆序數？
一個數組的逆序數表示將其轉換為已排序形式所需的更改次數。當一個數組已經排序時，它需要 0 個逆序，而在其他情況下，如果陣列倒置，則逆序數的數量將會最大。
為了解決這個問題，我們將遵循歸併排序方法以減少時間複雜度，並將其用於分治法演算法。
輸入
A sequence of numbers. (1, 5, 6, 4, 20).
輸出
將數字按升序排列所需的逆序數。
Here the number of inversions are 2. First inversion: (1, 5, 4, 6, 20) Second inversion: (1, 4, 5, 6, 20)
演算法
merge(array, tempArray, left, mid, right)
輸入 - 兩個陣列，進行了合併，左、右和中間索引。
輸出 - 已排序的合併陣列。
Begin i := left, j := mid, k := right count := 0 while i <= mid -1 and j <= right, do if array[i] <= array[j], then tempArray[k] := array[i] increase i and k by 1 else tempArray[k] := array[j] increase j and k by 1 count := count + (mid - i) done while left part of the array has some extra element, do tempArray[k] := array[i] increase i and k by 1 done while right part of the array has some extra element, do tempArray[k] := array[j] increase j and k by 1 done return count End
mergeSort(array, tempArray, left, right)
輸入 - 給定陣列和臨時陣列，陣列的左和右索引。
輸出 - 排序後的逆序數數量。
Begin count := 0 if right > left, then mid := (right + left)/2 count := mergeSort(array, tempArray, left, mid) count := count + mergeSort(array, tempArray, mid+1, right) count := count + merge(array, tempArray, left, mid+1, right) return count End
示例
#include <iostream>
using namespace std;
int merge(int arr[], int temp[], int left, int mid, int right) {
int i, j, k;
int count = 0;
i = left; //i to locate first array location
j = mid; //i to locate second array location
k = left; //i to locate merged array location
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]){ //when left item is less than right item
temp[k++] = arr[i++];
} else {
temp[k++] = arr[j++];
count += (mid - i); //find how many convertion is performed
}
}
while (i <= mid - 1) //if first list has remaining item, add them in the list
temp[k++] = arr[i++];
while (j <= right) //if second list has remaining item, add them in the list
temp[k++] = arr[j++];
for (i=left; i <= right; i++)
arr[i] = temp[i]; //store temp Array to main array
return count;
}
int mergeSort(int arr[], int temp[], int left, int right){
int mid, count = 0;
if (right > left) {
mid = (right + left)/2; //find mid index of the array
count = mergeSort(arr, temp, left, mid); //merge sort left sub array
count += mergeSort(arr, temp, mid+1, right); //merge sort right sub array
count += merge(arr, temp, left, mid+1, right); //merge two sub arrays
}
return count;
}
int arrInversion(int arr[], int n) {
int temp[n];
return mergeSort(arr, temp, 0, n - 1);
}
int main() {
int arr[] = {1, 5, 6, 4, 20};
int n = 5;
cout << "Number of inversions are "<< arrInversion(arr, n);
}輸出
Number of inversions are 2
更新於：30-7-2019
704 次瀏覽
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