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.
similar with max range sum.
class Solution {
public:
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
int n = gas.size();
vector<int> v(n);
for (int i = 0; i < n; i++) v[i] = gas[i] - cost[i];
int s = 0, e = 0, t = 0;
while (s < n) {
while (t >= 0 && e < s + n) t += v[(e++) % n];
if (t >= 0 && e == s + n) return s;
t -= v[s];
s++;
}
return -1;
}
};

本文介绍了一个算法,用于解决环形油站路线问题。给定一系列油站和油箱容量限制,该算法确定从哪个油站出发可以完成环形路线,确保油箱在行驶过程中不会空油。

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