Leetcode_1518_Water Bottles

[lihe]

Water Bottles

Given numBottles full water bottles, you can exchange numExchange empty water bottles for one full water bottle.

The operation of drinking a full water bottle turns it into an empty bottle.

Return the maximum number of water bottles you can drink.

Example 1:

Input: numBottles = 9, numExchange = 3
Output: 13
Explanation: You can exchange 3 empty bottles to get 1 full water bottle.
Number of water bottles you can drink: 9 + 3 + 1 = 13.

Example 2:

Input: numBottles = 15, numExchange = 4
Output: 19
Explanation: You can exchange 4 empty bottles to get 1 full water bottle. 
Number of water bottles you can drink: 15 + 3 + 1 = 19.

Example 3:

Input: numBottles = 5, numExchange = 5
Output: 6

Example 4:

Input: numBottles = 2, numExchange = 3
Output: 2

Constraints:

• 1 <= numBottles <= 100
• 2 <= numExchange <= 100

算法思路：

• exchange为交换来的新酒数量，初始化为0
• rest为剩下的空瓶的数量，初始化为numBottle + exchange
• rest = rest % numExchange + exchange
• 当rest < numExchange时结束循环

class Solution {
    public int numWaterBottles(int numBottles, int numExchange) {
    	int sum = 0;
    	sum += numBottles;
    	int exchange = 0;

    	int rest = numBottles + exchange;

    	while (rest >= numExchange){

    		exchange = rest / numExchange;
    		rest = rest % numExchange + exchange;
    		sum += exchange;

    	}
    	return sum;
    }
}