leetcode-gas station

Gas Station

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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.

Have you been asked this question in an interview?

最开始的方法，超时了：

class Solution {
public:
int canCompleteCircuit(vector<int> &gas, vector<int> &cost)
{
int len = gas.size();
int tank ;
int start,i;
for(start=0;start<len;start++)
{
tank = 0;
for(i=0;i<len;i++)
{
if(i+start<len)
{
tank += gas[i+start];
tank -= cost[i+start];
if(tank<0)break;
}
else
{
tank += gas[i+start-len];
tank -= cost[i+start-len];
if(tank<0)break;
}

}
if(i!=len)continue;
else
break;
}
if(start==len)return -1;
return start;
}
};

int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
// Note: The Solution object is instantiated only once.
int total = 0;
int currentgas = 0;
int startpoint = -1;
int sz = gas.size();
for(int i = 0; i < sz; i++)
{
currentgas += gas[i] - cost[i];
total += gas[i] - cost[i];
if(currentgas < 0)
{
startpoint = i;
currentgas = 0;
}
}
return total >= 0 ? startpoint+1 : -1;
}

解题报告：

当然，这题如是一个一个开始，然后把所有点走一遍，再判断是否成立，这个是可以的，也是最正常的思维，但复杂度O(n2)

这就是第一种解法

不得不优化。

有如下规律，如果一串数字，总和大于等于0，则必存在某一点，从这点开始，一真累加，每次都是非负数

这样，我们只需要判断最后的结果是否为非负就知道是否存在解，即total>=0

要找到开始的那个点，就判断是否是一直>=0，