js动态规划---最少硬币找零问题

给定钱币的面值 1、5、10、25

需要找给客户 36

最少找零数为: 1、10、25

 // dp[0] = 0 金额为零时不需要硬币

// dp[n] = min(dp[n],dp[n-coin1] + 1,dp[n-coin2],...) 金额为n时,硬币数等于(n-coin)+1中所需硬币最少的组合

function coinChange (coins,amount){
	let dp = new Array(amount+1).fill(Infinity);
	dp[0] = 0
	let path = [];
	for(let i = 1; i <= amount; i++){
		for(let j = 0; j < coins.length; j++){
			let n = i - coins[j];
			if(n >= 0){
				if(dp[n]+1  < dp[i]){
					dp[i] = dp[n]+1;
					path[i] = coins[j];
				}
			}
		}
	}
	
	let sum = path[amount];
	let res = [sum];
	amount = amount - sum;
	while(amount > 0){
		sum = path[amount];
		res.push(sum);
		amount = amount - sum;
	}
	
	
	console.log(dp,res);
}


coinChange([1,5,10,25],37);

  

原文地址:https://www.cnblogs.com/muamaker/p/9284350.html