标签：algorithm
There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of
gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.
Return the starting gas station‘s index if you can travel around the circuit once, otherwise return -1.
Note:
The solution is guaranteed to be unique.
class Solution {
public:
// pass every station,the cost is gas[i] - cost[i]
// if all the station cost sum is negative, then you can not pass a circle. otherwise, you can
// start from some place
// From one station, step by step, at one station if the cost sum become negative,
// then you must not start from this station, neighter the station before whose cost is bigger than zero
// so the time complexity is o(n)
//7:25
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
int len = gas.size();
int sum = 0,i=0,res=0;
vector<int> remain(len,0);<span style="white-space:pre"> </span>// calculate the single gas ramining in every station
for(i=0;i<len;i++)
{
remain[i] = gas[i] - cost[i];
sum += remain[i];
}
if(sum<0)
{
return -1;
}
sum = 0;<span style="white-space:pre"> </span>// suppose start from the zero
for(i=0;i<len;i++)
{
sum+=remain[i];<span style="white-space:pre"> </span> // use the front stations as the start place is not acceptable, may be start
<span style="white-space:pre"> </span> // from next station.
if(sum<0){
res = i+1;
sum = 0;
}
}
return res;
}
};标签：algorithm
原文地址：http://blog.csdn.net/nan327347465/article/details/39156887
