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 (0,2 are the person‘s ID),the transactions can be represented as [[0,10],[2,5]]
.
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,z). Note that
x ≠ y
andz > 0
. - Person‘s IDs may not be linear,e.g. we Could have the persons 0,2 or we Could also have the persons 0,2,6.
Example 1:
Input:
[[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,[1,1],5],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.
Runtime 24 ms,faster than 30.39%
这题本来以为是图,结果是数组的题。先建立每一个人的一个账户,然后DFS,DFS的时候从0-n,把第i个账户的钱都转到某一个j,这两个账户的金额是相反的,
因为只有相反的才会有交易。
class Solution { public: int minTransfers(vector<vector<int>>& transactions) { map<int,int> m; for(auto t: transactions){ m[t[0]] -= t[2]; m[t[1]] += t[2]; } vector<int> accnt(m.size()); int cnt = 0; for(auto a : m){ if(a.second != 0) accnt[cnt++] = a.second; } return helper(accnt,0,cnt,0); } int helper(vector<int>& accnt,int start,int n,int num){ int ret = INT_MAX; while(start < n && accnt[start] == 0) start++; for(int i=start+1; i<n; i++){ if((accnt[start] < 0 && accnt[i] > 0) || (accnt[start] > 0 && accnt[i] < 0)){ accnt[i] += accnt[start];//加入第i个账户中,这个时候start账户已经没有钱了,可以进行下一个账户清理了 ret = min(ret,helper(accnt,start+1,n,num+1));//DFS,保存最小值 accnt[i] -= accnt[start]; } } return ret == INT_MAX ? num : ret; } };
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