Gas Station.txt

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.
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思路：第一反应是以每个gas为起点进行遍历，注意第二层循环的数组index取(i+index)%len。复杂度为O(n2)
更好的O(n)解法：建立两个变量sum和total，从0到len遍历gas[]和cost[]，sum和total都等于gas[i]-cost[i]，检测sum是否小于0，
如果是，res变量等于当前index，同时sum归零。最终根据total变量是否大于零决定返回res或者-1。

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int sum = 0;
        int total = 0;
        int idx = -1;
        for(int i = 0; i < gas.size(); ++i)
        {
            sum += gas[i] - cost[i];
            total += gas[i] - cost[i];
            if(sum < 0)
            {
                idx = i;
                sum = 0;
            }
        }
        return total >= 0 ? idx + 1 : -1;
    }
};