Leetcode | | 134, gas station

Problem:

There was N gas stations along a circular route, where the amount of gas at Station I was gas[i] .

You had a car with an unlimited gas tank and it costs of gas to travel from station cost[i] I to its next station (i+ 1). You begin the journey with a 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.

Hide TagsGreedyTest instructions: The car from a gas station, refueling, consumption, find the only one can return to the original point of the gas station

Thinking:

(1) The algorithm is relatively simple, gasoline allowance is negative when not feasible, greedy strategy

(2) I have just started to adopt the dfs+ scissor branching way, the submission timed out, but this idea applies more broadly

(3) The use of array processing to avoid the cost of recursive function calls

Code

dfs+ Shear Branch: Timeout

Class Solution {Public:int cancompletecircuit (vector<int>& gas, vector<int>& cost) {int de        P=0;        int Maxdep=gas.size ()-1;        if (maxdep<0) return 0;        int fuel=0;        BOOL Flag=true;            for (int index=0;index<gas.size (); index++) {DFS (dep,maxdep,index,fuel,gas,cost,flag);            if (flag) return index;        Flag=true;    } return-1; }protected:void dfs (int dep,int maxdep,int index,int fuel, vector<int> &gas, vector<int> &cost, bo        Ol &flag) {if (!flag) return;        Fuel+=gas[index];        Fuel-=cost[index];            if (fuel<0) {flag=false;        Return            } if (DEP==MAXDEP) {if (fuel<0) Flag=false;            else flag=true;        Return    } dfs (DEP+1,MAXDEP, (index+1)%gas.size (), Fuel,gas,cost,flag); }    };

Greedy strategy: AC

Time complexity O (n)

Class Solution {public:    int cancompletecircuit (vector<int> &gas, vector<int> &cost) {        int Total = 0;        Int j =-1;        for (int i = 0, sum = 0; i < gas.size (); ++i) {            sum + = gas[i]-cost[i];            Total + = Gas[i]-cost[i];            if (Sum < 0) {                j = i;                sum = 0;            }        }        Return Total >= 0? J + 1:-1;    

Leetcode | | 134, gas station