gas-station

红太狼 2021-03-29 13:57 369阅读 0赞

There are *N* gas stations along a circular route, where the amount of gas at station *i* isgas\[i\].

You have a car with an unlimited gas tank and it costscost\[i\]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.

//题目求解，求解最大diff=gas\[i\]-cost\[i\]的最大值。贪心算法，如果sum<0 从该节点重新开始计算。

class Solution {
    public:
        int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
            if(gas.size()==0 || cost.size()==0 || gas.size()!=cost.size()) return 0;
            
            int total = 0,sum = 0,start = 0;
            
            for(int i = 0;i < gas.size();i++){        
                total += gas[i]-cost[i];
                if(sum < 0){
                    sum = gas[i]-cost[i];
                    start = i;
                }
                else{
                    sum += gas[i]-cost[i];
                }   
            }
            return total<0?-1:start;
        }
    };

gas-station

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