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 <iostream>

#include <stdio.h>

#include <vector>

using namespace std;

 

// 时间复杂度 O(n)，空间复杂度 O(1)

classSolution {

public:

    intcanCompleteCircuit(vector<int> &gas, vector<int> &cost) {

        inttotal = 0;

        intj = -1;

        for(inti = 0, sum = 0; i < gas.size(); ++i) {

            sum += gas[i] - cost[i];

            total += gas[i] - cost[i];

            if(sum < 0) {

                j = i;

                sum = 0;

            }

        }

        returntotal >= 0? j + 1: -1;

    }

};

 

intmain() {

    Solution solution;

    intresult;

    vector<int> gas =  {0,4,5};

    vector<int> cost = {1,2,6};

    result = solution.canCompleteCircuit(gas,cost);

    printf("Result:%d\n",result);

    return0;

}