134. Gas Station (Medium)

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: 9ms

这道转圈加油问题不算很难，只要想通其中的原理就很简单。我们首先要知道能走完整个环的前提是gas的总量要大于cost的总量，这样才会有起点的存在。假设开始设置起点start = 0, 并从这里出发，如果当前的gas值大于cost值，就可以继续前进，此时到下一个站点，剩余的gas加上当前的gas再减去cost，看是否大于0，若大于0，则继续前进。当到达某一站点时，若这个值小于0了，则说明从起点到这个点中间的任何一个点都不能作为起点，则把起点设为下一个点，继续遍历。当遍历完整个环时，当前保存的起点即为所求。代码如下：

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. (Should I prove them?) Here is my solution based on those ideas:

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

Proof:

1. 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.

2. Proof for the second point:

   1. If there is only one gas station, it’s true.
   2. 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.
   3. 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:
   4. either if gas(b) < cost(b), i.e., we cannot travel from b to c directly, then gas(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.
   5. 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.
   6. 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.