【leetcode】Gas Station

Gas Station

链接：https://oj.leetcode.com/problems/gas-station/

描述：

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.
方法一：
暴力求解。超时 O(n^2)。

代码如下：

int canCompleteCircuit(vector<int> &gas, vector<int> &cost)
{
int size = gas.size();
if( size < 0 ) return -1;

for(int i=0; i < size; ++i)
{
int totalgas = 0;
int totalcost = 0;
int num = 0;
while( num < size )
{
totalgas += gas[i + num];
totalcost += cost[i + num];
if(totalgas < totalcost)
break;
num++;
}
if( num == size)
return i;
}
return -1;
}

方法二：

每一次不符合要求时，充分借助前面的计算信息：

如果在A不能B(第一个不能到）说明A和B之间的任意一个都不能到B，因此下一次测试从B开始，一次类推。

如果totalgas 大于等于 totalcost，说明可以环绕。

第一条反证法很容易证明。至于第二条暂时没有好的方法证明，逆反定理是否可以。

代码如下：

int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
int start = 0;
int total = 0;
int tank = 0;

for(int i=0; i < gas.size(); ++i)
{
tank = tank + gas[i] - cost[i];
if( tank < 0)
{
start = i + 1;
total += tank;
tank = 0;
}
}

return (total + tank)>= 0 ? start : -1;
}