LeetCode (Gas Station)

题目要求：

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.

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，idx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。

代码：

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost)
    {
        int sum = 0, total = 0;
        int idx = -1;
        for (size_t i = 0; i < gas.size(); ++i) {
            int diff = gas[i] - cost[i];
            sum += diff;
            if(sum < 0)
            {
                sum = 0;
                idx = i;
            }
            total += diff;
        }
        return total >= 0 ? idx + 1: -1;//idx保存了使sum变为负数的index，所以要从下一个开始。
    }
};