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:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int length=gas.size();
vector<int> tmpGas(length,0);
for(int i=0;i<length;++i)
{
tmpGas[i]=gas[i]-cost[i];//兩者之差組成的新數組,代表補充或消耗
}
int startIndex=0;//初始起點
int sum=tmpGas[startIndex];
int i=0;
while(true)
{
i++;
if(i>=length)
{
i=i-length;
}
if(sum<0)
{
startIndex=i;//以下一個節點作爲起點
if(startIndex==0)
return -1;//又回到初始起點,開始出現循環
sum=tmpGas[startIndex];
}
else
{
sum+=tmpGas[i];
<span style="white-space:pre"> </span>
//已經到達終點,判斷是否能夠順利開完全程
if(startIndex==0 && i==length-1)
{//如果起點是0,當前結點又是最後一個節點
if(sum<0)
return -1;
else
return startIndex;
}
if(i==startIndex-1)
{//如果當前結點是循環的最後一個結點
if(sum>=0)
return startIndex;
else
return -1;
}
}
}
}
};