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73 lines
2 KiB
Kotlin
73 lines
2 KiB
Kotlin
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/**
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* File: coin_change.kt
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* Created Time: 2024-01-25
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* Author: curtishd (1023632660@qq.com)
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*/
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package chapter_dynamic_programming
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import java.util.*
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import kotlin.math.min
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/* 零钱兑换:动态规划 */
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fun coinChangeDP(coins: IntArray, amt: Int): Int {
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val n = coins.size
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val MAX = amt + 1
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// 初始化 dp 表
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val dp = Array(n + 1) { IntArray(amt + 1) }
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// 状态转移:首行首列
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for (a in 1..amt) {
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dp[0][a] = MAX
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}
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// 状态转移:其余行和列
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for (i in 1..n) {
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for (a in 1..amt) {
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if (coins[i - 1] > a) {
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// 若超过目标金额,则不选硬币 i
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dp[i][a] = dp[i - 1][a]
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} else {
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// 不选和选硬币 i 这两种方案的较小值
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dp[i][a] = min(dp[i - 1][a].toDouble(), (dp[i][a - coins[i - 1]] + 1).toDouble())
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.toInt()
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}
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}
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}
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return if (dp[n][amt] != MAX) dp[n][amt] else -1
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}
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/* 零钱兑换:空间优化后的动态规划 */
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fun coinChangeDPComp(coins: IntArray, amt: Int): Int {
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val n = coins.size
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val MAX = amt + 1
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// 初始化 dp 表
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val dp = IntArray(amt + 1)
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Arrays.fill(dp, MAX)
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dp[0] = 0
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// 状态转移
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for (i in 1..n) {
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for (a in 1..amt) {
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if (coins[i - 1] > a) {
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// 若超过目标金额,则不选硬币 i
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dp[a] = dp[a]
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} else {
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// 不选和选硬币 i 这两种方案的较小值
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dp[a] = min(dp[a].toDouble(), (dp[a - coins[i - 1]] + 1).toDouble()).toInt()
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}
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}
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}
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return if (dp[amt] != MAX) dp[amt] else -1
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}
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/* Driver Code */
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fun main() {
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val coins = intArrayOf(1, 2, 5)
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val amt = 4
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// 动态规划
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var res = coinChangeDP(coins, amt)
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println("凑到目标金额所需的最少硬币数量为 $res")
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// 空间优化后的动态规划
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res = coinChangeDPComp(coins, amt)
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println("凑到目标金额所需的最少硬币数量为 $res")
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}
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