题目描述：

设计一个存取款接口，入参是账户数组balances 与存取款请求体数组requests
对于取款要求判断：
当前余额不足，返回余额不足帐号
之前的取款时间在24之前的，在24小时之后返回上次取款额度的百分之2并向下取整。

example1:
String[] requests = {“withdraw 1613327630 2 480”,
“withdraw 1613327644 2 800”,
“withdraw 1614105244 1 100”,
“deposit 1614108844 2 200”,
“withdraw 1614108845 2 150”};
int[] balances = {1000, 1500};
返回： ｛900， 295｝

example2:
String[] requests = {“withdraw 1613327630 2 480”,
“withdraw 1613327644 2 800”,
“withdraw 1614105244 1 1000”,
“deposit 1614108844 2 200”,
“withdraw 1614108845 2 150”};
int[] balances = {1000, 500};
返回 ｛-1｝
注明：1号账户余额不足

思路

we can use an array preWithdrawNum to present the last withdraw account nums, each time we deal with the current request, we can judge whether the current time bigger than the last withdraw request 's time, if so, we can deal with the cashback logic of giving 2 percent of the last withdraw amount

 static int[] solution(int[] balances, String[] requests) {int[] preWithdrawNum = new int[requests.length];Arrays.fill(preWithdrawNum, -1);for (int i = 0; i < requests.length; i++) {String req = requests[i];String[] items = req.split(" ");String func = items[0];Integer num = Integer.valueOf(items[2]) - 1, amount = Integer.valueOf(items[3]);long time = Long.valueOf(items[1]);if (items.length != 4) {return new int[]{-num};}for (int j = 0; j < preWithdrawNum.length; j++) {if (preWithdrawNum[j] != -1 ) {String[] lastItems = requests[j].split(" ");long lastTimePlus24h = Long.valueOf(lastItems[1]) + 24 * 60 * 60;Integer currentAccountN = preWithdrawNum[j] ;if (time > lastTimePlus24h ) {balances[currentAccountN] += Math.floor(Integer.valueOf(lastItems[3]) * 0.02);preWithdrawNum[j] = -1;}}}if ("withdraw".equals(func)) {if (balances[num] < amount) {return new int[]{-i};}balances[num] -= amount;preWithdrawNum[i] = num;}if ("deposit".equals(func)) {balances[num] += amount;}}return balances;}