Feature/chapter dynamic programming swift (#608)

* feat: add Swift codes for intro_to_dynamic_programming article * feat: add Swift codes for dp_problem_features article * feat: add Swift codes for dp_solution_pipeline article * feat: add Swift codes for knapsack_problem article * feat: add Swift codes for unbounded_knapsack_problem article * feat: add Swift codes for edit_distance_problem article
2025-11-02 12:58:42 +08:00 · 2023-07-18 12:49:03 +08:00
parent 34985bdf2b
commit 9ea8a73059
13 changed files with 855 additions and 0 deletions
--- a/codes/swift/chapter_dynamic_programming/coin_change.swift
+++ b/codes/swift/chapter_dynamic_programming/coin_change.swift
@ -0,0 +1,69 @@
+/**
+ * File: coin_change.swift
+ * Created Time: 2023-07-15
+ * Author: nuomi1 (nuomi1@qq.com)
+ */
+
+/* 零钱兑换：动态规划 */
+func coinChangeDP(coins: [Int], amt: Int) -> Int {
+    let n = coins.count
+    let MAX = amt + 1
+    // 初始化 dp 表
+    var dp = Array(repeating: Array(repeating: 0, count: amt + 1), count: n + 1)
+    // 状态转移：首行首列
+    for a in stride(from: 1, through: amt, by: 1) {
+        dp[0][a] = MAX
+    }
+    // 状态转移：其余行列
+    for i in stride(from: 1, through: n, by: 1) {
+        for a in stride(from: 1, through: amt, by: 1) {
+            if coins[i - 1] > a {
+                // 若超过背包容量，则不选硬币 i
+                dp[i][a] = dp[i - 1][a]
+            } else {
+                // 不选和选硬币 i 这两种方案的较小值
+                dp[i][a] = min(dp[i - 1][a], dp[i][a - coins[i - 1]] + 1)
+            }
+        }
+    }
+    return dp[n][amt] != MAX ? dp[n][amt] : -1
+}
+
+/* 零钱兑换：状态压缩后的动态规划 */
+func coinChangeDPComp(coins: [Int], amt: Int) -> Int {
+    let n = coins.count
+    let MAX = amt + 1
+    // 初始化 dp 表
+    var dp = Array(repeating: MAX, count: amt + 1)
+    dp[0] = 0
+    // 状态转移
+    for i in stride(from: 1, through: n, by: 1) {
+        for a in stride(from: 1, through: amt, by: 1) {
+            if coins[i - 1] > a {
+                // 若超过背包容量，则不选硬币 i
+                dp[a] = dp[a]
+            } else {
+                // 不选和选硬币 i 这两种方案的较小值
+                dp[a] = min(dp[a], dp[a - coins[i - 1]] + 1)
+            }
+        }
+    }
+    return dp[amt] != MAX ? dp[amt] : -1
+}
+
+@main
+enum CoinChange {
+    /* Driver Code */
+    static func main() {
+        let coins = [1, 2, 5]
+        let amt = 4
+
+        // 动态规划
+        var res = coinChangeDP(coins: coins, amt: amt)
+        print("凑到目标金额所需的最少硬币数量为 \(res)")
+
+        // 状态压缩后的动态规划
+        res = coinChangeDPComp(coins: coins, amt: amt)
+        print("凑到目标金额所需的最少硬币数量为 \(res)")
+    }
+}