1 题目描述

环形路上有n个加油站，第i个加油站的汽油量是gas[i].
你有一辆车，车的油箱能够无限装汽油。从加油站i走到下一个加油站（i+1）花费的油量是cost[i]，你从一个加油站出发，刚开始的时候油箱里面没有汽油。
求从哪一个加油站出发能够在环形路上走一圈。返回加油站的下标，若是没有答案的话返回-1。
注意：
答案保证惟一。web

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.svg

示例1

输入ui

[2, 3, 1], [3, 1, 2]spa

输出code

1xml

2 解题思路

情景应用题，比较简单，直接看代码吧。token

3 代码实现

class Solution {
public:
    /** * * @param gas int整型vector * @param cost int整型vector * @return int整型 */
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        // write code here
        int numOfStations = gas.size();
        for(int i = 0; i < numOfStations; i++){
            int tank = 0;
            for(int j = 0; j < numOfStations; j++){
                int k = (i + j) % numOfStations;
                tank += gas[k] - cost[k];
                if(tank < 0)
                    break;
            }
            if(tank >= 0)
                return i;
        }
        return -1;
    }
};

4 运行结果

运行时间: 3ms
占用内存: 480KB内存