2021.1.14 刷题(零钱兑换-动规)

题目链接:https://leetcode-cn.com/problems/coin-change
题目描述:
给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额,返回 -1。
你可以认为每种硬币的数量是无限的。
示例 1:
输入:coins = [1, 2, 5], amount = 11
输出:3
解释:11 = 5 + 5 + 1

示例 2:
输入:coins = [2], amount = 3
输出:-1

示例 3:
输入:coins = [1], amount = 0
输出:0

示例 4:
输入:coins = [1], amount = 1
输出:1

示例 5:
输入:coins = [1], amount = 2
输出:2

提示:
1 <= coins.length <= 12
1 <= coins[i] <= 231 - 1
0 <= amount <= 104

动态规划问题讲解:
https://blog.csdn.net/zw6161080123/article/details/80639932
https://labuladong.gitbook.io/algo/dong-tai-gui-hua-xi-lie/dong-tai-gui-hua-ji-ben-ji-qiao/dong-tai-gui-hua-xiang-jie-jin-jie

代码:

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int N = coins.size();
        //动态二维数组定义,初始值为amount+1
         vector<vector<long>> dp(N + 1, vector<long>(amount + 1, amount + 1));
        //base case
        int i, j;
        //金额为0时,找钱数量为0
        for (i = 0; i <= N; i++) {
            dp[i][0] = 0;
        }
        //迭代
        for (i = 1; i <= N; i++) {
            for (j = 1; j <= amount; j++) {
                dp[i][j] = dp[i - 1][j]; 
                if (j >= coins[i - 1]) {
                    dp[i][j] = min(dp[i][j], dp[i][j - coins[i - 1]] + 1 );
                }
            }
        }

        
        if(dp[N][amount] == amount + 1)
        {
             return -1;
        }
           
        return dp[N][ amount];
    }
};
posted @ 2021-01-14 16:13  张宵  阅读(58)  评论(0编辑  收藏  举报