This commit is contained in:
krahets
2024-05-06 14:40:36 +08:00
parent 7e7eb6047a
commit 5c7d2c7f17
54 changed files with 3456 additions and 215 deletions
@@ -73,9 +73,41 @@ We have created an `Item` class in order to sort the items by their unit value.
=== "C++"
```cpp title="fractional_knapsack.cpp"
[class]{Item}-[func]{}
/* Item */
class Item {
public:
int w; // Item weight
int v; // Item value
[class]{}-[func]{fractionalKnapsack}
Item(int w, int v) : w(w), v(v) {
}
};
/* Fractional knapsack: Greedy */
double fractionalKnapsack(vector<int> &wgt, vector<int> &val, int cap) {
// Create an item list, containing two properties: weight, value
vector<Item> items;
for (int i = 0; i < wgt.size(); i++) {
items.push_back(Item(wgt[i], val[i]));
}
// Sort by unit value item.v / item.w from high to low
sort(items.begin(), items.end(), [](Item &a, Item &b) { return (double)a.v / a.w > (double)b.v / b.w; });
// Loop for greedy selection
double res = 0;
for (auto &item : items) {
if (item.w <= cap) {
// If the remaining capacity is sufficient, put the entire item into the knapsack
res += item.v;
cap -= item.w;
} else {
// If the remaining capacity is insufficient, put part of the item into the knapsack
res += (double)item.v / item.w * cap;
// No remaining capacity left, thus break the loop
break;
}
}
return res;
}
```
=== "Java"
+18 -1
View File
@@ -48,7 +48,24 @@ The implementation code is as follows:
=== "C++"
```cpp title="coin_change_greedy.cpp"
[class]{}-[func]{coinChangeGreedy}
/* Coin change: Greedy */
int coinChangeGreedy(vector<int> &coins, int amt) {
// Assume coins list is ordered
int i = coins.size() - 1;
int count = 0;
// Loop for greedy selection until no remaining amount
while (amt > 0) {
// Find the smallest coin close to and less than the remaining amount
while (i > 0 && coins[i] > amt) {
i--;
}
// Choose coins[i]
amt -= coins[i];
count++;
}
// If no feasible solution is found, return -1
return amt == 0 ? count : -1;
}
```
=== "Java"
+20 -1
View File
@@ -117,7 +117,26 @@ The variables $i$, $j$, and $res$ use a constant amount of extra space, **thus t
=== "C++"
```cpp title="max_capacity.cpp"
[class]{}-[func]{maxCapacity}
/* Maximum capacity: Greedy */
int maxCapacity(vector<int> &ht) {
// Initialize i, j, making them split the array at both ends
int i = 0, j = ht.size() - 1;
// Initial maximum capacity is 0
int res = 0;
// Loop for greedy selection until the two boards meet
while (i < j) {
// Update maximum capacity
int cap = min(ht[i], ht[j]) * (j - i);
res = max(res, cap);
// Move the shorter board inward
if (ht[i] < ht[j]) {
i++;
} else {
j--;
}
}
return res;
}
```
=== "Java"
@@ -6,7 +6,7 @@ comments: true
!!! question
Given a positive integer $n$, split it into at least two positive integers that sum up to $n$, and find the maximum product of these integers, as illustrated below.
Given a positive integer $n$, split it into at least two positive integers that sum up to $n$, and find the maximum product of these integers, as illustrated in Figure 15-13.
![Definition of the maximum product cutting problem](max_product_cutting_problem.assets/max_product_cutting_definition.png){ class="animation-figure" }
@@ -96,7 +96,26 @@ Please note, for the boundary case where $n \leq 3$, a $1$ must be split out, wi
=== "C++"
```cpp title="max_product_cutting.cpp"
[class]{}-[func]{maxProductCutting}
/* Maximum product of cutting: Greedy */
int maxProductCutting(int n) {
// When n <= 3, must cut out a 1
if (n <= 3) {
return 1 * (n - 1);
}
// Greedy cut out 3s, a is the number of 3s, b is the remainder
int a = n / 3;
int b = n % 3;
if (b == 1) {
// When the remainder is 1, convert a pair of 1 * 3 into 2 * 2
return (int)pow(3, a - 1) * 2 * 2;
}
if (b == 2) {
// When the remainder is 2, do nothing
return (int)pow(3, a) * 2;
}
// When the remainder is 0, do nothing
return (int)pow(3, a);
}
```
=== "Java"