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Update 0518.零钱兑换II.md

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java版本二维dp数组代码逻辑优化, 原版本的二维dp数组和一维dp数组的逻辑并不对应.
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EmmIons
2024-01-20 18:02:03 +08:00
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commit af40568275
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problems/0518.零钱兑换II.md
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@ -224,20 +224,26 @@ class Solution {
// 二维dp数组版本，方便理解
class Solution {
public int change(int amount, int[] coins) {
int[][] dp = new int[coins.length][amount + 1];
// 只有一种硬币的情况
for (int i = 0; i <= amount; i += coins[0]) {
dp[0][i] = 1;
int[][] dp = new int[coins.length][amount+1];
// 初始化边界值
for(int i = 0; i < coins.length; i++){
// 第一列的初始值为1
dp[i][0] = 1;
}
for (int i = 1; i < coins.length; i++) {
for (int j = 0; j <= amount; j++) {
// 第i种硬币使用0~k次，求和
for (int k = 0; k * coins[i] <= j; k++) {
dp[i][j] += dp[i - 1][j - k * coins[i]];
}
for(int j = coins[0]; j <= amount; j++){
// 初始化第一行
dp[0][j] += dp[0][j-coins[0]];
}
for(int i = 1; i < coins.length; i++){
for(int j = 1; j <= amount; j++){
if(j < coins[i]) dp[i][j] = dp[i-1][j];
else dp[i][j] = dp[i][j-coins[i]] + dp[i-1][j];
}
}
return dp[coins.length - 1][amount];
return dp[coins.length-1][amount];
}
}
```