Gas Station

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.

分析：一开始以为可以正方向和反方向进行讨论，通过几次提交发现，原来只需要考虑正方向，而且每组输入只有一个解！

以下为错误解法。

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        if(gas.size() == 1) return gas[0] >= cost[0] ? 0 : -1;
        
        for(int i = 0; i < gas.size(); i++){
            if(positive(gas, cost, i) || reverse(gas, cost, i)) return i;
        }
        return -1;
    }
    bool positive(vector<int> gas, vector<int> cost, int k){
        int currentGas = 0;
        for(int i = k; i < gas.size() + k; i++){
            int index = i % gas.size();
            currentGas += gas[index] - cost[index];
            if(currentGas < 0) return false;
        }
        return true;
    }
    bool reverse(vector<int> gas, vector<int> cost, int k){
        int currentGas = 0;
        for(int i = k; i > k - gas.size(); i--){
            int index = (i + gas.size()) % gas.size();
            int index1 = (i - 1 + gas.size()) % gas.size();
            currentGas += gas[index] - cost[index1];
            if(currentGas < 0) return false;
        }
        return true;
    }
};

由于如果能从结点i到结点j，那么从i+1可能能到j，可能不能到j。但是如果从结点i不能到结点j，那么从结点i+1之后的点都不能到j，因为i~i+1这一段路油箱至少会有剩余，否则不能从i开到i+1。所以，如果从点i开始计算的话，那么下一个该计算的点就是j。

运行时间：8ms

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int current = 0, whole = 0, result = 0;
        for(int i = 0; i < gas.size(); i++){
            current += gas[i] - cost[i];
            whole += gas[i] - cost[i]; //to judge whether result is valid
            if(current < 0){
                current = 0;
                result = i + 1;
            }
        }
        return whole < 0 ? -1 : result;
    }
};