LeetCode Gas Station 数学
作者:互联网
There are n
gas stations along a circular route, where the amount of gas
at the \(i\)th station is gas[i]
.
You have a car with an unlimited gas tank and it costs cost[i]
of gas to travel from the \(i\)th station to its next \((i + 1)\)th station. You begin the journey with an empty tank at one of the gas stations.
Given two integer arrays gas
and cost
, return the starting gas
station's index if you can travel around the circuit once in the clockwise direction, otherwise return \(-1\). If there exists a solution, it is guaranteed to be unique
Solution
给定一个圆环,有若干个加油站,问是否存在一个起始点使得空油箱的汽车跑完一周。不妨记:
\[dif[i] = gas[i]-cost[i] \]假设从 \(i\) 开始,此时到达 \(i+1\) 需要满足:
\[res[i+1] = gas[i]-cost[i]=dif[i]\geq 0 \]类似地,到达 \(i+2\) 需要满足:
\[dif[i]+dif[i+1]\geq 0 \]可以发现这就是前缀和,对于环形问题,我们可以摊开为一个 \(2n\) 的序列即可。所以只需要所有的前缀和非负即可
点击查看代码
class Solution {
private:
int dif[200005];
int ans = 0;
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int n = gas.size();
int ck = 0;
for(int i=0;i<n;i++)dif[i] = gas[i]-cost[i], ck += dif[i];
for(int i=n;i<2*n;i++)dif[i] = dif[i-n];
int fg=0;
if(ck<0)return -1;
int tank_V = 0, st = 0;
for(int i=0;i<2*n;i++){
tank_V += dif[i];
if(tank_V<0){
tank_V = 0;
st = i+1;
}
}
return st%n;
}
};
标签:cost,int,Gas,dif,gas,Station,th,station,LeetCode 来源: https://www.cnblogs.com/xinyu04/p/16479492.html