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Copy pathCoinChange2.cpp
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84 lines (76 loc) · 2.26 KB
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// Source : https://leetcode.com/problems/coin-change-2/
// Author : Hao Chen
// Date : 2019-03-18
/*****************************************************************************************************
*
* You are given coins of different denominations and a total amount of money. Write a function to
* compute the number of combinations that make up that amount. You may assume that you have infinite
* number of each kind of coin.
*
* Example 1:
*
* Input: amount = 5, coins = [1, 2, 5]
* Output: 4
* Explanation: there are four ways to make up the amount:
* 5=5
* 5=2+2+1
* 5=2+1+1+1
* 5=1+1+1+1+1
*
* Example 2:
*
* Input: amount = 3, coins = [2]
* Output: 0
* Explanation: the amount of 3 cannot be made up just with coins of 2.
*
* Example 3:
*
* Input: amount = 10, coins = [10]
* Output: 1
*
* Note:
*
* You can assume that
*
* 0 <= amount <= 5000
* 1 <= coin <= 5000
* the number of coins is less than 500
* the answer is guaranteed to fit into signed 32-bit integer
*
******************************************************************************************************/
class Solution {
public:
int change(int amount, vector<int>& coins) {
return change_dp(amount, coins);
return change_recursive(amount, coins); // Time Limit Error
}
int change_recursive(int amount, vector<int>& coins) {
int result = 0;
change_recursive_helper(amount, coins, 0, result);
return result;
}
// the `idx` is used for remove the duplicated solutions.
void change_recursive_helper(int amount, vector<int>& coins, int idx, int& result) {
if (amount == 0) {
result++;
return;
}
for ( int i = idx; i < coins.size(); i++ ) {
if (amount < coins[i]) continue;
change_recursive_helper(amount - coins[i], coins, i, result);
}
return;
}
int change_dp(int amount, vector<int>& coins) {
vector<int> dp(amount+1, 0);
dp[0] = 1;
for (int i=0; i<coins.size(); i++) {
for(int n=1; n<=amount; n++) {
if (n >= coins[i]) {
dp[n] += dp[n-coins[i]];
}
}
}
return dp[amount];
}
};