There are numBottles water bottles that are initially full of water. You can exchange numExchange empty water bottles from the market with one full water bottle.

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

Given the two integers numBottles and numExchange, 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.

Constraints:
1 <= numBottles <= 100
2 <= numExchange <= 100

換水問題。

超市正在促銷，你可以用 numExchange 個空水瓶從超市兌換一瓶水。最開始，你一共購入了 numBottles 瓶水。

如果喝掉了水瓶中的水，那么水瓶就會變成空的。

給你兩個整數(shù) numBottles 和 numExchange ，返回你 最多 可以喝到多少瓶水。

思路

這是一道模擬題。題目說了每 numExchange 個空瓶可以換一瓶水，那么我們可以模擬這個過程，每喝掉 numExchange 瓶水，就去兌換一瓶水，直到 numBottles 喝光。注意兌換的水要加回 numBottles。

復(fù)雜度

時間O(n)
空間O(1)

代碼

Java實現(xiàn)

class Solution {
    public int numWaterBottles(int numBottles, int numExchange) {
        int count = 0;
        while (numBottles >= numExchange) {
            count += numExchange;
            numBottles -= numExchange - 1;
        }
        return count + numBottles;
    }
}