Description:

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.

Code:

    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
if (gas.size()== || cost.size()== || gas.size()!=cost.size() )
return -;
int total = ;
int sum = ;
int startStation = ;
for (int i = ; i < gas.size(); ++i)
{
int diff = (gas[i]-cost[i]);
total += diff;
if (sum < )
{
sum = diff;
startStation = i;
}
else
sum += diff;
}
return total < ?-:startStation;
}

判断条件:

从0号加油站开始,找到能开到最后一个加油站的加油站q,并且全部加油站的加油量减去消耗量的指非负(即能从0号加油站开到q),则q为我们要找的加油站。

分析:

这道题最直观的思路,是逐个尝试每一个站点,从站 i 点出发,看看是否能走完全程。如果不行,就接着试着从站点 i+1出发。

假设从站点 i 出发,到达站点 k 之前,依然能保证油箱里油没见底儿,从k 出发后,见底儿了。那么就说明 diff[i] + diff[i+1] + ... + diff[k] < 0,而除掉diff[k]以外,从diff[i]开始的累加都是 >= 0的。也就是说diff[i] 也是 >= 0的,这个时候我们还有必要从站点 i + 1 尝试吗?仔细一想就知道:车要是从站点 i+1出发,到达站点k后,甚至还没到站点k,油箱就见底儿了,因为少加了站点 i 的油。。。

因此,当我们发现到达k 站点邮箱见底儿后,i 到 k 这些站点都不用作为出发点来试验了,肯定不满足条件,只需要从k+1站点尝试即可!因此解法时间复杂度从O(n2)降到了 O(2n)。之所以是O(2n),是因为将k+1站作为始发站,车得绕圈开回k,来验证k+1是否满足。

等等,真的需要这样吗?

我们模拟一下过程:

a. 最开始,站点0是始发站,假设车开出站点p后,油箱空了,假设sum1 = diff[0] +diff[1] + ... + diff[p],可知sum1 < 0;

b. 根据上面的论述,我们将p+1作为始发站,开出q站后,油箱又空了,设sum2 = diff[p+1] +diff[p+2] + ... + diff[q],可知sum2 < 0。

c. 将q+1作为始发站,假设一直开到了未循环的最末站,油箱没见底儿,设sum3 = diff[q+1] +diff[q+2] + ... + diff[size-1],可知sum3 >= 0。

要想知道车能否开回 q 站,其实就是在sum3 的基础上,依次加上 diff[0] 到 diff[q],看看sum3在这个过程中是否会小于0。但是我们之前已经知道 diff[0] 到 diff[p-1] 这段路,油箱能一直保持非负,因此我们只要算算sum3 + sum1是否 <0,就知道能不能开到 p+1站了。如果能从p+1站开出,只要算算sum3 + sum1 + sum2 是否 < 0,就知都能不能开回q站了。

因为 sum1, sum2 都 < 0,因此如果 sum3 + sum1 + sum2 >=0 那么 sum3 + sum1 必然 >= 0,也就是说,只要sum3 + sum1 + sum2 >=0,车必然能开回q站。而sum3 + sum1 + sum2 其实就是 diff数组的总和 Total,遍历完所有元素已经算出来了。因此 Total 能否 >= 0,就是是否存在这样的站点的 充分必要条件

这样时间复杂度进一步从O(2n)降到了 O(n)。

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