1.题目要求:

示例:

输入：
["Bank", "withdraw", "transfer", "deposit", "transfer", "withdraw"]
[[[10, 100, 20, 50, 30]], [3, 10], [5, 1, 20], [5, 20], [3, 4, 15], [10, 50]]
输出：
[null, true, true, true, false, false]解释：
Bank bank = new Bank([10, 100, 20, 50, 30]);
bank.withdraw(3, 10);    // 返回 true ，账户 3 的余额是 $20 ，所以可以取款 $10 。// 账户 3 余额为 $20 - $10 = $10 。
bank.transfer(5, 1, 20); // 返回 true ，账户 5 的余额是 $30 ，所以可以转账 $20 。// 账户 5 的余额为 $30 - $20 = $10 ，账户 1 的余额为 $10 + $20 = $30 。
bank.deposit(5, 20);     // 返回 true ，可以向账户 5 存款 $20 。// 账户 5 的余额为 $10 + $20 = $30 。
bank.transfer(3, 4, 15); // 返回 false ，账户 3 的当前余额是 $10 。// 所以无法转账 $15 。
bank.withdraw(10, 50);   // 返回 false ，交易无效，因为账户 10 并不存在。提示：n == balance.length
1 <= n, account, account1, account2 <= 105
0 <= balance[i], money <= 1012
transfer, deposit, withdraw 三个函数，每个 最多调用 104 次

2.题目代码：

class Bank {
public:vector<long long> bank_account;Bank(vector<long long>& balance) {bank_account = balance;}//转帐bool transfer(int account1, int account2, long long money) {if(account1 > bank_account.size() || account2 > bank_account.size()){return false;}if(bank_account[account1 - 1] >= money){bank_account[account1 - 1] -= money;bank_account[account2 - 1] += money;return true;}else{return false;}}//存款bool deposit(int account, long long money) {if(account > bank_account.size()){return false;}else{bank_account[account - 1] += money;return true;}}//取款bool withdraw(int account, long long money) {if(account > bank_account.size()){return false;}if(bank_account[account - 1]  >=  money){bank_account[account - 1] -= money;return true;}else{return false;}}
};/*** Your Bank object will be instantiated and called as such:* Bank* obj = new Bank(balance);* bool param_1 = obj->transfer(account1,account2,money);* bool param_2 = obj->deposit(account,money);* bool param_3 = obj->withdraw(account,money);*/