[LeetCode] Gas Station 加油站问题
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.
这道转圈加油问题不算很难,只要想通其中的原理就很简单。我们首先要知道能走完整个环的前提是gas的总量要大于cost的总量,这样才会有起点的存在。假设开始设置起点start = 0, 并从这里出发,如果当前的gas值大于cost值,就可以继续前进,此时到下一个站点,剩余的gas加上当前的gas再减去cost,看是否大于0,若大于0,则继续前进。当到达某一站点时,若这个值小于0了,则说明从起点到这个点中间的任何一个点都不能作为起点,则把起点设为下一个点,继续遍历。当遍历完整个环时,当前保存的起点即为所求。代码如下:
解法一:
class Solution { public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int total = 0, sum = 0, start = 0; for (int i = 0; i < gas.size(); ++i) { total += gas[i] - cost[i]; sum += gas[i] - cost[i]; if (sum < 0) { start = i + 1; sum = 0; } } return (total < 0) ? -1 : start; } };
我们也可以从后往前遍历,用一个变量mx来记录出现过的剩余油量的最大值,total记录当前剩余油量的值,start还是记录起点的位置。当total大于mx的时候,说明当前位置可以作为起点,更新start,并且更新mx。为啥呢?因为我们每次total加上的都是当前位置的油量减去消耗,如果这个差值大于0的话,说明当前位置可以当作起点,因为从当前位置到末尾都不会出现油量不够的情况,而一旦差值小于0的话,说明当前位置如果是起点的话,油量就不够,无法走完全程,所以我们不更新起点位置start。最后结束后我们还是看totoa是否大于等于0,如果其小于0的话,说明没有任何一个起点能走完全程,因为总油量都不够,参见代码如下:
解法二:
class Solution { public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int total = 0, mx = -1, start = 0; for (int i = gas.size() - 1; i >= 0; --i) { total += gas[i] - cost[i]; if (total > mx) { start = i; mx = total; } } return (total < 0) ? -1 : start; } };
类似题目:
Cheapest Flights Within K Stops
参考资料:
https://leetcode.com/problems/gas-station/discuss/42568/Share-some-of-my-ideas.
https://leetcode.com/problems/gas-station/discuss/42656/8ms-simple-O(n)-c++-solution