LeetCode #465 - Optimal Account Balancing
题目描述：
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:
1. A transaction will be given as a tuple (x, y, z). Note that x ≠ y and z > 0.
2. 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.
class Solution {
public:
int minTransfers(vector>& transactions) {
unordered_map hash; // 表示每个人账户上有多少钱
for (auto t:transactions)
{
hash[t[0]]-=t[2];
hash[t[1]]+=t[2];
}
vector account(hash.size());
int i=0;
for (auto x:hash)
{
if (x.second!=0)
{
account[i]=x.second; // 有多少个账户是不为零的
i++;
}
}
return helper(account,0,i,0);
}
int helper(vector& account, int start, int n, int num) // num是转账次数
{
int res=INT_MAX;
while(start0)||(account[i]>0&&account[start]<0))
{
account[i]+=account[start];
res=min(res,helper(account,start+1,n,num+1));
account[i]-=account[start];
}
}
if(res==INT_MAX) return num;
else return res;
}
};