算法课第五周 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.

环形路线上有N个加油站，每个加油站有汽油gas[i]，从每个加油站到下一站消耗汽油cost[i]，问从哪个加油站出发能够回到起始点，如果都不能则返回-1（注意，解是唯一的）。

代码 

#include
#include
#include
using namespace std;

// 时间复杂度 O(n)，空间复杂度 O(1)
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 ;
    }
};

int main() {
    Solution solution;
    int result;
    vector< int > gas =  { 0 , 4 , 5 };
    vector< int > cost = { 1 , 2 , 6 };
    result = solution.canCompleteCircuit(gas,cost);
    printf( "Result:%d\n" ,result);
    return 0 ;
}