LeetCode134：Gas Station

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.

解题思路：

一开始，我打算通过两层循环，依次从某个点出发并测试是否能够运行一圈，可想而知时间复杂度为O（n2），不满足要求。之后看了/article/1347625.html这篇博客的解题思路，才发现有更简单更优雅的解决方案，大概思路如下：

1 如果总的gas - cost小于零的话，那么没有解返回-1

2 如果前面所有的gas - cost加起来小于零，那么前面所有的点都不能作为出发点。

关于第一点无需多言，这里详解下第二点，为什么前面所有的点都不能作为起始站了，原因是：

假设从第0个站点开始，0~1，剩余的煤气left1 = gas[i]-cost[i],如果left为负，则过不去，必须从下一个站点从新开始，如果为正，则1~2时，left2 = gas[1]+left – cost[1],然后是2~3等等继续下去，如果left一直为正，则表示这些站点都可以过去，但当某个站点i~i+1时，left为负数，说明过不去，且之前的所有站点都不能作为起始站，因为，每个站点要到下一个站点时，gas = gas +left,都不能过去，现在如果从某个站点开始，即gas量为它自身，更过不去。

实现代码：

#include <iostream>
#include <vector>
using namespace std;

/*
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.
*/
class Solution {
public:
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
if(gas.size() == 0 || cost.size() == 0 || gas.size() != cost.size())
return -1;
int left = 0;
int total = 0;
int start = 0;
int n = gas.size();
for(int i = 0; i < n; i++)
{
left += gas[i] - cost[i];//从i到i+i，剩余的煤气
total += gas[i] - cost[i];
if(left < 0)//表示前面那些站点都不能作为起始站，现在开始从下一个站点开始
{
start = i + 1;
left = 0;
}
}
return total >= 0? start : -1;//煤气总量是否大于等于总消耗

}
};
ing main(void)
{
return 0;
}