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Merge Sort
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#include <iostream> | |
#include <vector> | |
using namespace std; | |
void merge(int l, int r, vector<int> &v) // Combina os resultados | |
{ | |
vector<int> S; // sequência auxiliar | |
int mid = (l + r) / 2; | |
int p1 = l; // aponta para o primeiro elemento da parte 1 | |
int p2 = mid + 1; // aponta para o primeiro elemento da parte 2 | |
while (p1 <= mid && p2 <= r) | |
if (v[p1] <= v[p2]) | |
{ | |
S.push_back(v[p1]); | |
p1++; | |
} | |
else | |
{ | |
S.push_back(v[p2]); | |
p2++; | |
} | |
while (p1 <= mid) // Se ainda sobrarem elementos na parte 1 | |
{ | |
S.push_back(v[p1]); | |
p1++; | |
} | |
while (p2 <= r) // Se ainda sobrarem elementos na parte 2 | |
{ | |
S.push_back(v[p2]); | |
p2++; | |
} | |
for (int i = 0; i < S.size(); i++) | |
v[l + i] = S[i]; | |
// agora, o intervalo [l, r] em v está ordenado | |
} | |
void mergeSort(int l, int r, vector<int> &v) // Divide o problema | |
{ | |
int mid; | |
if (l < r) | |
{ | |
mid = (l + r) / 2; | |
mergeSort(l, mid, v); // Ordena a parte 1 | |
mergeSort(mid + 1, r, v); // Ordena a parte 2 | |
merge(l, r, v); | |
} | |
} | |
int main() | |
{ | |
vector<int> x = {9, 4, 2, 1, -3, 1, 7, 19, 32, 23, -17}; | |
int N = (int)x.size(); | |
mergeSort(0, N - 1, x); | |
return 0; | |
} |
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def merge(l, r, v): | |
S = [] | |
mid = (l+r) // 2 | |
p1 = l | |
p2 = mid + 1 | |
while p1 <= mid and p2 <= r: | |
if (v[p1] <= v[p2]): | |
S.append(v[p1]) | |
p1 += 1 | |
else: | |
S.append(v[p2]) | |
p2 += 1 | |
while p1 <= mid: | |
S.append(v[p1]) | |
p1 += 1 | |
while p2 <= r: | |
S.append(v[p2]) | |
p2 += 1 | |
for i in range(len(S)): | |
v[l + i] = S[i] | |
def mergeSort(l, r, v): | |
if (l < r): | |
mid = (l+r) // 2 | |
mergeSort(l, mid, v) | |
mergeSort(mid+1, r, v) | |
merge(l, r, v) | |
def main(): | |
x = [9, 4, 2, 1, -3, 1, 7, 19, 32, 23, -17] | |
N = len(x) | |
mergeSort(0, N-1, x) | |
if __name__ == '__main__': | |
main() |
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