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coinChangeII.java
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60 lines (58 loc) · 2.06 KB
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import java.util.*;
public class coinChangeII {
public static void main(String[] args){
int[] coins = new int[] {1,2,5};
int amount = 11;
map = new HashMap<>();
map.put(Arrays.asList(1,3), 0);
List<Integer> key = Arrays.asList(1,3);
System.out.println(map.containsKey(key));
System.out.println(dfs(coins, 0, amount));
}
/* DP solution:
Think about this question from very easy sub-problem: if we only use coin 1, how can we build that
then we add coin 2.. coin 5..
int[] coins = [1,2,5]
* */
public static int coinChange(int[] coins, int amount){
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int coin : coins) {
for (int i = coin; i <= amount; i++) {
dp[i] += dp[i-coin];
}
}
return dp[amount];
}
/* 折腾半天终于写出了memorized DFS solution
* */
// List<index, amount>; Integer: count
// 用来做memorized dfs的map
// 注意要用list！不要用int[]！！！
static Map<List<Integer>, Integer> map;
public static int dfs(int[] coins, int index, int amount){
// map里面有可以直接返回
List<Integer> key = Arrays.asList(index, amount);
if(map.containsKey(key)){
System.out.println(key);
return map.get(key);
}
// base case
if (amount <= 0) {
if(amount == 0){
return 1;
}
return 0;
}
int cnt = 0;
for(int i = index; i < coins.length; i++){
// 终于理解了 应该memorize 什么了
// 观察每次调用dfs的时候，是index和amount两个参数在变化（int[] coins一直不变）。
// 当index和amount与之前的相同时，说明曾经出现过，我们做了相同的work
// 理解：重复解决子问题：用index之后的coins，去满足这个amount
cnt += dfs(coins, i, amount-coins[i]);
}
map.put(key, cnt);
return cnt;
}
}