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e720aa2d24
* Sync recent changes to the revised Word. * Revised the preface chapter * Revised the introduction chapter * Revised the computation complexity chapter * Revised the chapter data structure * Revised the chapter array and linked list * Revised the chapter stack and queue * Revised the chapter hashing * Revised the chapter tree * Revised the chapter heap * Revised the chapter graph * Revised the chapter searching * Reivised the sorting chapter * Revised the divide and conquer chapter * Revised the chapter backtacking * Revised the DP chapter * Revised the greedy chapter * Revised the appendix chapter * Revised the preface chapter doubly * Revised the figures
70 lines
2 KiB
C++
70 lines
2 KiB
C++
/**
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* File: coin_change.cpp
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* Created Time: 2023-07-11
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* Author: Krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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/* 零钱兑换：动态规划 */
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int coinChangeDP(vector<int> &coins, int amt) {
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int n = coins.size();
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int MAX = amt + 1;
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// 初始化 dp 表
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vector<vector<int>> dp(n + 1, vector<int>(amt + 1, 0));
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// 状态转移：首行首列
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for (int a = 1; a <= amt; a++) {
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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 (int i = 1; i <= n; i++) {
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for (int a = 1; a <= amt; a++) {
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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], dp[i][a - coins[i - 1]] + 1);
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}
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}
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}
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return dp[n][amt] != MAX ? dp[n][amt] : -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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int coinChangeDPComp(vector<int> &coins, int amt) {
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int n = coins.size();
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int MAX = amt + 1;
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// 初始化 dp 表
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vector<int> dp(amt + 1, MAX);
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dp[0] = 0;
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// 状态转移
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for (int i = 1; i <= n; i++) {
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for (int a = 1; a <= amt; a++) {
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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], dp[a - coins[i - 1]] + 1);
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}
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}
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}
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return dp[amt] != MAX ? dp[amt] : -1;
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}
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/* Driver code */
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int main() {
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vector<int> coins = {1, 2, 5};
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int amt = 4;
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// 动态规划
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int res = coinChangeDP(coins, amt);
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cout << "凑到目标金额所需的最少硬币数量为 " << res << endl;
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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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cout << "凑到目标金额所需的最少硬币数量为 " << res << endl;
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return 0;
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}
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