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.
Subscribe to see which companies asked this question
题意是有N个加油站点构成一个环状,每个站点i加油量为gas[i],从站点i到站点i+1需要耗费油量为cost[i],现要求从哪个站点出发可以成功转一圈回到初始站点,返回该站点,若没有则返回-1。
totalGas=gas[0]+…+gas[len];
totalCost=cost[0]+…+cost[len];
如果成功那么totalGas-totalCost>=0。
remain表示从开始点start到当前点i剩下的油量,如果为负,说明start~i这段中没有符合条件的点作为开始点,所以从i+1又开始检查。
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
if(gas.size()==0 || cost.size()==0){
return -1;
}
int len=gas.size();
int remain=0;
int start=0;
int total=0;
for(int i=0;i<len;i++){
total+=(gas[i]-cost[i]);
if(remain<0){
remain=gas[i]-cost[i];
start=i;
}
else{
remain+=(gas[i]-cost[i]);
}
}
return total<0?-1:start;
}
};