The fifth week of algorithm class gas station

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 cost[i] of the gas-to-travel from station 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.

There are n petrol stations on the ring route, petrol at each petrol station gas[i], from each gas station to the next station consuming petrol cost[i], ask from which gas station to go back to the starting point, if neither can return-1 (note that the solution is unique).

Code

#include <iostream> #include <stdio.h> #include <vector> using namespace std;  //Time complexity O (n), Space complexity O (1) class Solution {public:      int cancompletecircuit (vector< int & Gt &gas, vector< int > &cost) {         int total = 0;   &n bsp;      Int J =-1;          for (int i = 0, sum = 0; i < gas.size (); ++i) {   & nbsp;         sum + = gas[i]-cost[i];              Total + = Gas[i]-cost[i];              if (Sum < 0) {    & nbsp;            j = i;                  sum = 0;