题目来源
一道简单的DP题,n种硬币,要求组成某个数值的硬币数最少,代码如下:
class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
vector<int> dp(amount+1, -1);
dp[0] = 0;
for (int i=1; i<amount+1; i++) {
for (int j=0; j<coins.size(); j++) {
if (i >= coins[j] && dp[i-coins[j]] != -1)
dp[i] = (dp[i] == -1) ? dp[i-coins[j]] + 1 : min(dp[i], dp[i-coins[j]] + 1);
}
}
return dp[amount];
}
};
看了下讨论区,感觉可以写的更简洁一下,代码如下:
class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
int Max = amount+1;
vector<int> dp(amount+1, Max);
dp[0] = 0;
for (int i=1; i<amount+1; i++) {
for (int j=0; j<coins.size(); j++) {
if (i >= coins[j])
dp[i] = min(dp[i], dp[i-coins[j]] + 1);
}
}
return dp[amount] > amount ? -1 : dp[amount];
}
};