A group of friends went on holiday and sometimes lent each other money. For example, Alice paid for Bill’s lunch for $10. Then later Chris gave Alice $5 for a taxi ride. We can model each transaction as a tuple (x, y, z) which means person x gave person y $z. Assuming Alice, Bill, and Chris are person 0, 1, and 2 respectively (0, 1, 2 are the person’s ID), the transactions can be represented as [[0, 1, 10], [2, 0, 5]].
Given a list of transactions between a group of people, return the minimum number of transactions required to settle the debt.
Note:
- A transaction will be given as a tuple (x, y, z). Note that
x ≠ yandz > 0. - Person’s IDs may not be linear, e.g. we could have the persons 0, 1, 2 or we could also have the persons 0, 2, 6.
Example 1:
Input:
[[0,1,10], [2,0,5]]
Output:
2
Explanation:
Person #0 gave person #1 $10.
Person #2 gave person #0 $5.
Two transactions are needed. One way to settle the debt is person #1 pays person #0 and #2 $5 each.
Example 2:
Input:
[[0,1,10], [1,0,1], [1,2,5], [2,0,5]]
Output:
1
Explanation:
Person #0 gave person #1 $10.
Person #1 gave person #0 $1.
Person #1 gave person #2 $5.
Person #2 gave person #0 $5.
Therefore, person #1 only need to give person #0 $4, and all debt is settled.
这道题给了一堆某人欠某人多少钱这样的账单，问经过优化后最少还剩几个。其实就相当于一堆人出去玩，某些人可能帮另一些人垫付过花费，最后结算总花费的时候可能你欠着别人的钱，其他人可能也欠你的欠，需要找出简单的方法把所有欠账都还清就行了。这道题的思路跟之前那道 Evaluate Division 有些像，都需要对一组数据颠倒顺序处理。这里使用一个 HashMap 来建立每个人和其账户的映射，其中账户若为正数，说明其他人欠你钱；如果账户为负数，说明你欠别人钱。对于每份账单，前面的人就在 HashMap 中减去钱数，后面的人在哈希表中加上钱数。这样每个人就都有一个账户了，接下来要做的就是合并账户，看最少需要多少次汇款。先统计出账户值不为0的人数，因为如果为0了，表明你既不欠别人钱，别人也不欠你钱，如果不为0，把钱数放入一个数组 accnt 中，然后调用递归函数。在递归函数中，首先跳过为0的账户，然后看若此时 start 已经是 accnt 数组的长度了，说明所有的账户已经检测完了，用 cnt 来更新结果 res。否则就开始遍历之后的账户，如果当前账户和之前账户的钱数正负不同的话，将前一个账户的钱数加到当前账户上，这很好理解，比如前一个账户钱数是 -5，表示张三欠了别人5块钱，当前账户钱数是5，表示某人欠了李四5块钱，那么张三给李四5块，这两人的账户就都清零了。然后调用递归函数，此时从当前改变过的账户开始找，cnt 表示当前的转账数，需要加1，后面别忘了复原当前账户的值，典型的递归写法，参见代码如下：
class Solution {
public:
int minTransfers(vector<vector<int>>& transactions) {
int res = INT_MAX;
unordered_map<int, int> m;
for (auto t : transactions) {
m[t[0]] -= t[2];
m[t[1]] += t[2];
}
vector<int> accnt;
for (auto a : m) {
if (a.second != 0) accnt.push_back(a.second);
}
helper(accnt, 0, 0, res);
return res;
}
void helper(vector<int>& accnt, int start, int cnt, int& res) {
int n = accnt.size();
while (start < n && accnt[start] == 0) ++start;
if (start == n) {
res = min(res, cnt);
return;
}
for (int i = start + 1; i < n; ++i) {
if ((accnt[i] < 0 && accnt[start] > 0) || (accnt[i] > 0 && accnt[start] < 0)) {
accnt[i] += accnt[start];
helper(accnt, start + 1, cnt + 1, res);
accnt[i] -= accnt[start];
}
}
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/465
类似题目：
参考资料：
https://leetcode.com/problems/optimal-account-balancing/
LeetCode All in One 题目讲解汇总(持续更新中…)
转载请注明来源于 Grandyang 的博客 (grandyang.com)，欢迎对文章中的引用来源进行考证，欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论，也可以邮件至 grandyang@qq.com
