【Gas Station】cpp

题目：

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.

代码：

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
            if ( gas.size() != cost.size() ) return -;
            int start_station = -;
            int tmp_sum = ;
            int total = ;
            for (size_t i = ; i < gas.size(); ++i)
            {
                tmp_sum += gas[i] - cost[i];
                total += gas[i] - cost[i];
                if (tmp_sum<)
                {
                    tmp_sum=;
                    start_station = i;
                }
            }
            return total>= ? start_station+ : -;
    }
};

Tips:

1. 采用贪心算法的思想：维护一个tmp_sum，计算从开始节点到当前节点损耗之差；如果小于零，则直接放弃；否则，继续累加。

2. 最终判断能不能完成行程，需要维护一个total：如果total大于等于0，则判定一定可以走完这趟旅程。这是为什么呢？具体原理可以参见鸽巢原理。

参考资料
贪心算法：http://zh.wikipedia.org/wiki/贪心法

鸽巢原理：http://zh.wikipedia.org/wiki/鴿巢原理

===========================================================

第二次过这道题，一次AC了，大体思路记得比较清楚。写法完全按照自己理解了，与原来的代码已经不太一样了。

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
            const int gas_n = gas.size();
            const int cost_n = cost.size();
            if ( gas_n!= cost_n ) return -;
            int total_compare = ;
            int curr_gas = ;
            int start = ;
            for ( int i=; i<cost_n; ++i )
            {
                if ( curr_gas== ) start = i;
                curr_gas = curr_gas + gas[i] - cost[i];
                if ( curr_gas< ) curr_gas = ;
                total_compare = total_compare + gas[i] - cost[i];
            }
            return total_compare>= ? start : -;
    }
};