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72 lines
1.9 KiB
C
72 lines
1.9 KiB
C
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/**
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* File: coin_change_ii.c
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* Created Time: 2023-10-02
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* Author: Zuoxun (845242523@qq.com)
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*/
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#include "../utils/common.h"
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/* 零钱兑换 II:动态规划 */
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int coinChangeIIDP(int coins[], int amt, int coinsSize) {
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int n = coinsSize;
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// 初始化 dp 表
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int dp[n + 1][amt + 1];
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memset(dp, 0, sizeof(dp));
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// 初始化首列
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for (int i = 0; i <= n; i++) {
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dp[i][0] = 1;
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}
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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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// 若超过背包容量,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 这两种方案之和
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dp[i][a] = dp[i - 1][a] + dp[i][a - coins[i - 1]];
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}
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}
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}
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return dp[n][amt];
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}
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/* 零钱兑换 II:空间优化后的动态规划 */
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int coinChangeIIDPComp(int coins[], int amt, int coinsSize) {
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int n = coinsSize;
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// 初始化 dp 表
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int dp[amt + 1];
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memset(dp, 0, sizeof(dp));
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dp[0] = 1;
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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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// 若超过背包容量,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 这两种方案之和
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dp[a] = dp[a] + dp[a - coins[i - 1]];
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}
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}
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}
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return dp[amt];
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}
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/* Driver code */
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int main() {
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int coins[] = {1, 2, 5};
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int coinsSize = sizeof(coins) / sizeof(coins[0]);
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int amt = 5;
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// 动态规划
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int res = coinChangeIIDP(coins, amt, coinsSize);
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printf("凑出目标金额的硬币组合数量为 %d\n", res);
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// 空间优化后的动态规划
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res = coinChangeIIDPComp(coins, amt, coinsSize);
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printf("凑出目标金额的硬币组合数量为 %d\n", res);
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return 0;
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}
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