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1e49574332
Use PascalCase for all structs in C. SImplify n_queens.c Format C code for chapter of graph.
60 lines
1.7 KiB
C
60 lines
1.7 KiB
C
/**
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* File: coin_change_greedy.c
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* Created Time: 2023-09-07
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* Author: lwbaptx (lwbaptx@gmail.com)
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*/
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#include "../utils/common.h"
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/* 零钱兑换:贪心 */
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int coinChangeGreedy(int *coins, int size, int amt) {
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// 假设 coins 列表有序
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int i = size - 1;
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int count = 0;
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// 循环进行贪心选择,直到无剩余金额
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while (amt > 0) {
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// 找到小于且最接近剩余金额的硬币
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while (i > 0 && coins[i] > amt) {
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i--;
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}
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// 选择 coins[i]
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amt -= coins[i];
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count++;
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}
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// 若未找到可行方案,则返回 -1
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return amt == 0 ? count : -1;
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}
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/* Driver Code */
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int main() {
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// 贪心:能够保证找到全局最优解
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int coins1[6] = {1, 5, 10, 20, 50, 100};
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int amt = 186;
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int res = coinChangeGreedy(coins1, 6, amt);
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printf("\ncoins = ");
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printArray(coins1, 6);
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printf("amt = %d\n", amt);
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printf("凑到 %d 所需的最少硬币数量为 %d\n", amt, res);
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// 贪心:无法保证找到全局最优解
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int coins2[3] = {1, 20, 50};
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amt = 60;
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res = coinChangeGreedy(coins2, 3, amt);
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printf("\ncoins = ");
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printArray(coins2, 3);
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printf("amt = %d\n", amt);
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printf("凑到 %d 所需的最少硬币数量为 %d\n", amt, res);
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printf("实际上需要的最少数量为 3 ,即 20 + 20 + 20\n");
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// 贪心:无法保证找到全局最优解
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int coins3[3] = {1, 49, 50};
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amt = 98;
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res = coinChangeGreedy(coins3, 3, amt);
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printf("\ncoins = ");
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printArray(coins3, 3);
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printf("amt = %d\n", amt);
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printf("凑到 %d 所需的最少硬币数量为 %d\n", amt, res);
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printf("实际上需要的最少数量为 2 ,即 49 + 49\n");
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return 0;
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}
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