由分治(Divide&Conquer)通向歸併排序(MergeSort)
一個求數組中的最大值的分治實現:
/*
* @Descripttion:
* @version:
* @Author: iDestro
* @Date: 2020-03-17 20:23:45
* @LastEditors: iDestro
* @LastEditTime: 2020-03-17 20:27:21
*/
#include <iostream>
using namespace std;
int max(int arr[], int l, int r) {
if (l == r) {
return arr[l];
}
int mid = l + ((r-l)>>2);
int a = max(arr, l, mid);
int b = max(arr, mid+1, r);
return a > b ? a : b;
}
int main() {
int arr[] = {1, 2, 3, 4};
cout << max(arr, 0, 3);
return 0;
}
歸併排序:
/*
* @Descripttion:
* @version:
* @Author: iDestro
* @Date: 2020-03-15 15:33:39
* @LastEditors: iDestro
* @LastEditTime: 2020-03-15 15:49:08
*/
#include <iostream>
using namespace std;
void merge(int arr[], int l, int r) {
int *arr_tmp = new int[r-l+1];
int mid = l + ((r-l)>>2), i = 0;
int p1 = l, p2 = mid+1;
while (p1 <= mid && p2 <= r) {
arr_tmp[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++];
}
while (p1 <= mid) arr_tmp[i++] = arr[p1++];
while (p2 <= r) arr_tmp[i++] = arr[p2++];
for (int i = 0; i < r-l+1; i++) {
arr[l+i] = arr_tmp[i];
}
}
void merge_sort(int arr[], int l, int r) {
if (l == r) {
return;
}
int mid = l + ((r-l)>>2);
merge_sort(arr, l, mid);
merge_sort(arr, mid+1, r);
merge(arr, l, r);
}
int main() {
int arr[] = {4, 3, 2, 1};
merge_sort(arr, 0, 3);
for (int i:arr) {
cout << i << " ";
}
return 0;
}
歸併排序思想的應用:
小和問題:在一個數組中,每一個數左邊比當前數小的數累加起來,叫做這個數組的小和,求一個數組的小和。
輸入:
1 3 4 2 5
輸出:
16
分析:
1左邊比1小的數,沒有;
3左邊比3小的數,1;
4左邊比4小的數,1、3;
2左邊比2小的數,1、3、4、2;
所以小和爲1+1+3+1+1+3+4+2=16
代碼實現:
- Solution[0]:
int small_sum(int arr[], int n) {
int res = 0;
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (arr[j] < arr[i]) {
res += arr[j];
}
}
}
return res;
}
- Solution[1]:
int merge(int arr[], int l, int r) {
int *arr_tmp = new int[r-l+1];
int mid = l+((r-l)>>2), i = 0, res = 0;
int p1 = l, p2 = mid+1;
while (p1 <= mid && p2 <= r) {
if (arr[p1] < arr[p2]) {
res += (r-p2+1)*arr[p1];
arr_tmp[i++] = arr[p1++];
} else {
arr_tmp[i++] = arr[p2++];
}
}
while (p1 <= mid) arr_tmp[i++] = arr[p1++];
while (p2 <= r) arr_tmp[i++] = arr[p2++];
for (int i = 0; i < r-l+1; i++) {
arr[l+i] = arr_tmp[i];
}
return res;
}
int small_sum(int arr[], int l, int r) {
if (l == r) {
return 0;
}
int mid = l + ((r-l)>>2);
return small_sum(arr, l, mid) +
small_sum(arr, mid+1, r) +
merge(arr, l, r);
}