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.
Solution:
The brute force solution costs O(n^2), I find a O(n) and one round solution. If gas[i] >= cost[i], then I mark it, and calcuate the gas in tank along next node, if it is less than zero, then restart search.
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
int curr_gas = ;
int idx = -;
int tmp_gas = ;
for(int i = ; i < gas.size(); i ++) {
curr_gas += gas[i] - cost[i];
if( idx != -)
tmp_gas += gas[i] - cost[i];
if(tmp_gas < ) {
idx = -;
tmp_gas = ;
}
if(gas[i] - cost[i] >= && idx == -) {
idx = i;
tmp_gas += gas[i] - cost[i];
}
}
if(curr_gas >= )
return idx;
else
return -;
}