为什么我的硬币兑换功能在包含在记忆缓存中时不起作用

Question

我正在尝试经典的硬币更改问题，我的以下代码工作正常，例如它返回正确的值 3，硬币组合为 [1， 2， 5] 和目标 11。但是，当我将记忆添加到递归调用时，它会得出不正确的答案？我在函数调用中做错了什么？

var coinChange = function(coins, amount, totalCoins = 0, cache = {}) {
    if (cache[amount] !== undefined) return cache[amount];
    if (amount === 0) {
        return totalCoins;
    } else if (0 > amount) {
        return Number.MAX_SAFE_INTEGER;
    } else {
        let minCalls = Number.MAX_SAFE_INTEGER;
        for (let i = 0; i < coins.length; i++) {
            let recursiveCall = coinChange(coins, amount - coins[i], totalCoins + 1, cache);
            minCalls = Math.min(minCalls, recursiveCall);
        }
        const returnVal = (minCalls === Number.MAX_SAFE_INTEGER) ? -1 : minCalls;
        return cache[amount] = returnVal;
    }
}
console.log(coinChange([1, 2, 5], 11)); // This ends up outputting 7!?!?!

Answer 1

您不应该将totalCoins作为递归调用的函数参数传递，因为这是您尝试计算的内容。相反，它应该按如下方式计算

var coinChange = function(coins, amount, cache = {}) {
    if (cache[amount] !== undefined) return cache[amount];
    if (amount === 0) {
        return 0;
    } else if (0 > amount) {
        return Number.MAX_SAFE_INTEGER;
    } else {
        let minCalls = Number.MAX_SAFE_INTEGER;
        for (let i = 0; i < coins.length; i++) {
            let recursiveCall = 1 + coinChange(coins, amount - coins[i], cache);
            minCalls = Math.min(minCalls, recursiveCall);
        }
        const returnVal = (minCalls === Number.MAX_SAFE_INTEGER) ? -1 : minCalls;
        return cache[amount] = returnVal;
    }
}
console.log(coinChange([1, 2, 5], 11)); // This outputs 3

注意此行

let recursiveCall = 1 + coinChange(coins, amount - coins[i], cache);