134. Gas Station
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.
思路就是 一次遍历,
如果一段路的remain < 0 就说明 如果能走完,那么开始的位置肯定不是在 当前i的前面,而是在后面。所以start = i + 1
最后的remain+debt能判断 能不能走完。
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost)
{
int start = 0;
int debt = 0;
int remain = 0;
for (int i = 0; i < gas.size(); i++)
{
remain += gas[i]-cost[i];
if (remain < 0)
{
debt += remain;
start = i+1;
remain = 0;
}
}
return (remain + debt >= 0) ? start : -1;
}
};
本文探讨了一道经典的算法问题——环路加油站问题。该问题要求找到一条环形路径上能够完成一圈旅行的起点加油站。通过一次遍历算法,确定是否能从某个起点完成整个环路的行驶,并给出实现代码。
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