Gas Station

There are N gas stations along a circular route, where the amount of gas at stationi is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from stationi 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.

这道题使用贪心算法来解决：

I have thought for a long time and got two ideas:

• If car starts at A and can not reach B. Any station between A and B can not reach B.(B is the first station that A can not reach.)
• If the total number of gas is bigger than the total number of cost. There must be a solution.

Proof to the first point: say there is a point C between A and B -- that is A can reach C but cannot reach B. Since A cannot reach B, the gas collected between A and B is short of the cost. Starting from A, at the time when the car reaches C, it brings in gas >= 0, and the car still cannot reach B. Thus if the car just starts from C, it definitely cannot reach B.

Proof for the second point:

• If there is only one gas station, it’s true.
• If there are two gas stations a and b, and gas(a) cannot afford cost(a), i.e., gas(a) < cost(a), then gas(b) must be greater than cost(b), i.e., gas(b) > cost(b), since gas(a) + gas(b) > cost(a) + cost(b); so there must be a way too.
• If there are three gas stations a, b, and c, where gas(a) < cost(a), i.e., we cannot travel from a to b directly, then:
  • either if gas(b) < cost(b), i.e., we cannot travel from b to c directly, then cost(c) > cost(c), so we can start at c and travel to a; since gas(b) < cost(b), gas(c) + gas(a) must be greater than cost(c) + cost(a), so we can continue traveling from a to b. Key Point: this can be considered as there is one station at c’ with gas(c’) = gas(c) + gas(a) and the cost from c’ to b is cost(c’) = cost(c) + cost(a), and the problem reduces to a problem with two stations. This in turn becomes the problem with two stations above.
  • or if gas(b) >= cost(b), we can travel from b to c directly. Similar to the case above, this problem can reduce to a problem with two stations b’ and a, where gas(b’) = gas(b) + gas(c) and cost(b’) = cost(b) + cost(c). Since gas(a) < cost(a), gas(b’) must be greater than cost(b’), so it’s solved too.
• For problems with more stations, we can reduce them in a similar way. In fact, as seen above for the example of three stations, the problem of two stations can also reduce to the initial problem with one station.

code:

class Solution {public:    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {       int sum =0,total = 0,index = 0;       for(int i(0);i<gas.size();++i)       {           sum = sum+gas[i]-cost[i];           if(sum<0)           {                           total +=sum;               sum=0;               index = i+1;           }       }       return total+sum>=0?index:-1;    }};