Gas Station

题目描述:
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.

分析:
考虑从每个点出发的可行性即可

代码:

int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
    if(gas.size()==1) {
        if(gas[0]>=cost[0]) return 0;
        return -1;
      }
        for(int i=0;i<gas.size();i++){
            int num=0;
            int ok=0;
            int j;
            for( j=0;j<gas.size();j++){
            int s=i+j;
            if(s>=gas.size()) s=s%gas.size();
                num+=gas[s];
                num-=cost[s];
                if(num>=0){
                ok++;
                if(ok==gas.size()) return i;
                }
                else{
                    break;
                }
            }


        }
        return -1;
    }
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值