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Main.java
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Main.java
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class Solution {
//Runtime: 2 ms, faster than 99.97%
//Memory Usage: 36.4 MB, less than 81.19%
public int change(int amount, int[] coins) {
int types = coins.length;
// LinkedList<> dp = new LinkedList(amount+1);
int [] dp = new int[amount+1];// 不记得怎么进行2维数组的初始化,只好进行空间压缩
// 无限背包问题,dp为types行,amount+1列
// 第i行j列表示:仅包含coins[0]~coins[i]这些面额的硬币时,总价值为j时的组合个数
// dp[i][j] = dp[i-1][j-coins[i]] + dp[i-1][j];
// dp[0][0] = 1
dp[0] = 1;
for (int t = 0; t < types; ++t) {
for (int v = coins[t]; v <= amount; ++v) {
// dp[t][v] = dp[t-1][v-coins[t]] + dp[t-1][v];
dp[v] = dp[v-coins[t]] + dp[v];
}
}
return dp[amount];
// 测试一下[1,2,5] target = 5
// 1 2 3 4 5
// 1 0 0 0 0 0 initial dp
// \|\|\|\|\|
// 1 1 1 1 1 1 只使用面额coins[0]时的组合数
// \ \|\|\|\|
// \|\|\|\|
// 1 1 2 2 3 3 只使用面额coins[0]和coins[1]时,4 = 1,1,1,1, 4 = 2,2, 4=1,2,1
// \ |
// \ |
// \ |
// \ |
// \|
// 4 使用coins[0],coins[1] 和coins[2]时的组合数
}
}
public class Main {
//public static int main(String argv) { 这里输入是个[]
public static void main(String [] args) {
int rtn;
Solution gua = new Solution();
int amount;
int [] coins = {1,2,5};
rtn = gua.change(amount = 5, coins);
//System.print.ln("expact value = 4, actual value = " + rtn);
System.out.println("expact value = 4, actual value = " + rtn);
}
}