134. Gas Station
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Problem:
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.
题意是有一段环形的路,路上有n个加油站,第i个加油站对应就有gas[i]那么多的汽油,然后你开着一个汽车,汽车的装油没有上限,而你从第i个加油站到i+1个加油站要用cost[i]那么多的汽油,然后你从0汽油开始出发,问从哪个加油站开始可以绕一圈这段路,如果不存在的话返回-1。
如果用O(n^2)的当然是一下子就想到,对每个点都进行一次循环判断。但是这样可能就会超时,有没有更好的方法呢?
其实假如从第一个站开始,经过前k个站,发现汽油不够了,这时候的话这几个站都可以不用考虑了,因为本来油就少了,去掉开头的站也是缺油的。所以这个时候就重新从第k+1个站开始考虑。然后还要用一个total值来保存整个旅程的汽油剩余量,假如小于0,表示是不存在的,可以返回-1。sum用来计算经过后的剩余量。
Code:(LeetCode运行6ms)
class Solution {public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int sum = 0, total = 0; int result = -1; for (int i = 0; i < gas.size(); i++) { sum += gas[i] - cost[i]; total += gas[i] - cost[i]; if (sum < 0) { sum = 0; result = i; } } if (total >= 0) { return result + 1; } else { return -1; } }};
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