Created
January 7, 2018 08:42
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322. Coin Change
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/** | |
* @param {number[]} coins | |
* @param {number} amount | |
* @return {number} | |
*/ | |
var coinChange = function(coins, amount) { | |
// Using the greedy method. (Finding the maximum benefit.) | |
// 1. First, we should select the maximum denomination. | |
// 2. then, we should select the second largest coin. | |
// 3. Finally, you can find the fewest number of coins. | |
// 考慮 | |
// 1. [1,2,6,4] 這種 coins 沒有排列 | |
// 2. 如果直接從最大的 coins 那來,但如果是 [2,3,5] 11 這種案例,直接用 5 來找,會變成無法整除。 | |
// 總結上面使用 greedy 的缺點,不用他。 | |
//let result = 0; | |
//coins.sort((a,b) => a > b); | |
//for (var i=coins.length-1; i>=0 ; i--){ | |
// let coinNum = Math.floor(amount/coins[i]); | |
// result += coinNum; | |
// amount -= coins[i] * coinNum; | |
// console.log(`coin:${coins[i]} ,coinNum:${coinNum}, amount:${amount}`); | |
//} | |
//return (amount === 0) ? result : -1; | |
// DP | |
// 從 sum 0 開始往上加 | |
// 每一次算好 sum 時就將要完成這個 sum 的總數所需的最小硬幣數存在 cache 中。 | |
// 每一次計算時,先用 sum - coin 去看看 cache 有沒有值 | |
// 有值就取出來用,如果沒有就代表無法用這個 coin 來處理。 | |
let cache = []; | |
let sum = 0; | |
cache[0] = 0; | |
if(amount === 0){ | |
return 0; | |
} | |
coins = coins.filter((val) => { | |
return (val <= amount) ? true : false; | |
}) | |
while (++sum <= amount){ | |
let min = -1; | |
for (var i=0;i< coins.length;i++){ | |
if(sum < coins[i]) continue; | |
if(cache[sum-coins[i]] !== undefined){ | |
let temp = cache[sum-coins[i]] + 1; | |
min = (min < 0) ? temp : Math.min(temp,min); | |
} | |
} | |
if(min !== -1){ | |
cache[sum] = min; | |
} | |
} | |
return cache[amount] ? cache[amount] : -1; | |
}; |
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