标签：leetcode   算法   

问题描述：

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.

Note:
The solution is guaranteed to be unique.

代码：

int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {  //C++
        if(gas.size()==0|| cost.size()==0||gas.size()!=cost.size())
            return -1;
            
        int size = gas.size();
        vector<int> left(size);
        for(int i=0; i<size; i++)
            left[i]=gas[i]-cost[i];
        
        int max=left[0],begin=0,temp=0,index=begin;
        for(int i=0; i<size; i++ ){
            temp += left[i];
            if(temp > max){
                max = temp;
                index = begin;
            }
            if(temp < 0){
                temp = 0;
                begin = i+1;
            }
        }
        if(temp >=0 && begin !=0){
            for(int i=0; i<size; i++){
                temp += left[i];
                if(temp > max){
                    max =temp;
                    index = begin;
                }
                if(temp <0)
                    break;
            }
        }
        temp = 0;
        for(int i=0; i<size; i++ )
        {
            temp +=left[(i+index)%size];
            if(temp <0)
                return -1;
        }
        return index;
    }

[leetcode]Gas Station

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/42686365