zoukankan      html  css  js  c++  java
  • LeetCode-Coin Change

    You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

    Example 1:
    coins = [1, 2, 5], amount = 11
    return 3 (11 = 5 + 5 + 1)

    Example 2:
    coins = [2], amount = 3
    return -1.

    Note:
    You may assume that you have an infinite number of each kind of coin.

    Credits:
    Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

    Analysis:

    If you can get i, then you can get i+(coin:coins).

    Solution:

     1 public class Solution {
     2     public int coinChange(int[] coins, int amount) {
     3         if (amount==0) return 0;
     4         
     5         int[] num = new int[amount+1];
     6         Arrays.fill(num,Integer.MAX_VALUE);
     7         num[0] = 0;
     8         
     9         for (int i=0;i<=amount;i++){
    10             if (num[i]==Integer.MAX_VALUE) continue;
    11             for (int coin : coins){
    12                 if (coin+i <= amount){
    13                     num[coin+i] =Math.min(num[coin+i],num[i]+1);
    14                 }
    15             }
    16         }
    17         
    18         return (num[amount]==Integer.MAX_VALUE) ? -1 : num[amount];
    19     }
    20 }
  • 相关阅读:
    正课day04
    正科day03
    正课day02
    正课day01
    预科day08
    Elasticsearch之-文档操作
    Elasticsearch之-映射管理
    Elasticsearch之-索引操作
    Elasticsearch之-倒排索引
    es安装官方,第三方插件
  • 原文地址:https://www.cnblogs.com/lishiblog/p/5764938.html
Copyright © 2011-2022 走看看