Leetcode 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.

此题题目意思是从环型加油站中选择一个加油站出发，使其能够回到起始点。

最基本的方法是暴力，用一个二重循环解决

本题考虑到汽车到达一个加油加上gas[i]的油，行走到下一个加油站时，剩下的油为gas[i]-cost[i]

首先判断所有的sum(gas[0..n))是否大于sum(cost[0..n))的和

 如果小于的话，说明不可能有剩余的油，不可能对每个环形加油站都走一遍，故返回-1

如果大于等于的话，说明存在可能走完所有环形加油站

　　累积每个加油站加油行走后剩余的油即sum(gas[i]-cost[i])，标记初始索引

　　如果sum < 0说明从标记索引不可能走到，故重新标记一下一个索引，重新计数知道循环结束，返回标记索引

int canCompleteCircuit(vector<int>& gas, vector<int> &cost){
    vector<int> leave(gas.size(),0);
    transform(gas.begin(),gas.end(),cost.begin(), leave.begin(),minus<int>());
    if(accumulate(leave.begin(),leave.end(),0) < 0) return -1;
    int sum = 0, index = 0;
    for(int i = 0; i < leave.size(); ++ i){
        sum+=leave[i];
        if(sum < 0) sum = 0,index = i+1;
    }
    return index;
}

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Leetcode Gas Station

原文：http://www.cnblogs.com/xiongqiangcs/p/3796643.html