| 31 | |
| 32 | // limit可以限制硬币的数量，比如{5000:0,1000:4,}就是没有5000的，1000的有四个 |
| 33 | var getMoney1 = function( number,option={}) { |
| 34 | if(!number){ |
| 35 | return 0; |
| 36 | } |
| 37 | const coins = [5000,1000,500,100,50] |
| 38 | const obj = coins.reduce((ret,a)=>{ |
| 39 | ret[a] = 0 |
| 40 | return ret |
| 41 | },{}) |
| 42 | |
| 43 | let dp = Array(number+1).fill(Infinity) |
| 44 | dp[0] = 0 |
| 45 | let selected = Array(number+1) |
| 46 | selected[0] = {...obj} |
| 47 | const limit = {...obj,...option} |
| 48 | for(const coin of coins){ |
| 49 | let count = limit[coin] |
| 50 | for(let i=number;i>=0;i--){ |
| 51 | if(dp[i]!==Infinity){ |
| 52 | // 编译count, 钱数没超过number |
| 53 | for(let j=1;j<=count &&i+j*coin<=number;j++){ |
| 54 | // 比如dp[100]，2块的硬币有3个 |
| 55 | // 遍历这三个硬币，可以从dp[98]+1, dp[96]+2,dp[94]+3 三个最小值决定 |
| 56 | if(dp[i]+j<dp[i+j*coin]){ |
| 57 | dp[i+j*coin] = dp[i]+j |
| 58 | selected[i+j*coin] = {...selected[i]} |
| 59 | selected[i+j*coin][coin]+=j |
| 60 | } |
| 61 | } |
| 62 | } |
| 63 | } |
| 64 | } |
| 65 | return dp[number]===Infinity?-1:selected[number] |
| 66 | } |
| 67 | console.log(getMoney(6200)) |
| 68 | |
| 69 | const option = { 5000:0,1000:7,100:5} |