Question

我正在尝试经典的coin change problem，下面的代码可以正常工作，例如，它返回正确的值为3，硬币组合为[1、2、5]，目标为11。但是，当我将备忘录添加到递归调用中，导致答案不正确？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

您不应将totalCoins作为函数参数进行递归调用，因为这就是您要计算的内容。相反，应按以下方式计算

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

请注意此行

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

备忘缓存中包含我的硬币找零功能为何不起作用？

1 个答案: