[cc150] 硬币问题

Given an infinite number of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent), find how many ways to represent n cents.

思路:

从最大面值的硬币开始分析,然后依次看更小面值的硬币。假设 n = 100, 所有 valid 排列组合中

num_quarters = 0 的是一类,

num_quarters = 1 的是一类,

。。。

num_quarters = 4 的是一类

num_quarters = 0 的子集中,num_dimes = 0的是一类,num_dimes = 1的是一类,。。。

按照这种思路,可以简单的用下面的代码实现:

public int makeChange(int amount) {
    int[] denoms = {25, 10, 5, 1};
    return makeChange(amount, denoms, 0);
}

public int makeChange(int amount, int[] denoms, int index) {
    if(index == denoms.length - 1)
        return 1;
    
    int ways = 0;
    for(int i = 0; i <= amount / denoms[index]; ++i){
        ways += makeChange(amount - i * denoms[index], denoms, index+1);
    }
    return ways;
}

上面的代码是正确的,但是不够efficient,因为存在重复运算,比如一共有60 cents时,子集A{num_quarters=2, num_dimes=0}和子集B{num_quarters=0, num_dimes=5}是相同的,计算了两次。为了避免这种情况,下面的代码进行了优化

int makeChange(int n){
    int[] denoms = {25, 10, 5, 1};
    int[][] map = new int[n+1][denoms.length];
    return makeChange(n, denoms, 0, map);
}

int makeChange(int amount, int[] denoms, int index, int[][] map){
    if(map[amount][index] > 0){
        return map[amount][index];
    }
    
    if(index >= denoms.length - 1)
        return 1;
    
    int denomAmount = denoms[index];
    int ways = 0;
    for(int i = 0; i * denomAmount <= amount; i++){
        int amountRemaining = amount - i * denomAmount;
        ways += makeChange(amountRemaining, denoms, index +1 , map);
    }
    
    map[amount][index] = ways;
    return ways;
}

 

posted @ 2014-06-09 05:11  门对夕阳  阅读(479)  评论(0编辑  收藏  举报