zoukankan      html  css  js  c++  java
  • leetcode 322: 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.

    题意:所需硬币的最小数目

    思路:

    例:coins = [1,2,5],amount = 11

    则所需硬币数ans[11] = min(ans[11-1], ans[11-2], ans[11-5]);

    注意初始状态为-1,即不存在;

     1 class Solution {
     2 public:
     3     int coinChange(vector<int>& coins, int amount) {
     4         if(amount==0)
     5             return 0;
     6         int len = coins.size();
     7         vector<int> v(amount+1,-1);
     8         v[0] = 0;
     9         for(int i=1;i<=amount;i++)
    10         {
    11             for(int j=0;j<len;j++)
    12             {
    13                 int tmp = i-coins[j];
    14                 if(tmp<0)
    15                     continue;
    16                 if(v[tmp]==-1)
    17                     continue;
    18                 if(v[i]==-1)
    19                     v[i] = v[tmp]+1;
    20                 else
    21                     v[i] = min(v[i],v[tmp]+1);
    22             }
    23         }
    24         return v[amount];
    25     }
    26 
    27 };
    View Code
  • 相关阅读:
    JS中的constructor与prototype
    HTTP状态码
    CSS HACK 及常见问题
    js常见怪异
    js深拷贝和浅拷贝
    浏览器渲染
    google全球地址大全
    从function前面的!想到的
    2048
    js判定IE
  • 原文地址:https://www.cnblogs.com/jsir2016bky/p/5781914.html
Copyright © 2011-2022 走看看