题目:
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.
class Solution {
public:
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
int n = gas.size();
int cur = 0;
int total = 0;
int res = -1;
for (int i = 0; i < n; i++) {
cur += gas[i] - cost[i];
total += gas[i] - cost[i];
if (cur < 0) {
res = i;
cur = 0;
}
}
return total >= 0 ? res + 1 : -1;
}
};
本文探讨了一个关于循环路线中油站的问题,即如何确定一辆充满无限油箱的车辆从哪个油站出发,可以完成一圈行驶而不耗尽油。通过分析给定的代码实现,解释了算法背后的逻辑和解决问题的方法。
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