leetcode-gas station

Gas Station

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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.

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最开始的方法，超时了：

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

int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {        // Note: The Solution object is instantiated only once.int total = 0;int currentgas = 0; int startpoint = -1;int sz = gas.size();for(int i = 0; i < sz; i++){currentgas += gas[i] - cost[i];total += gas[i] - cost[i];if(currentgas < 0){startpoint = i;currentgas = 0;}}return total >= 0 ? startpoint+1 : -1;    }

解题报告：
当然，这题如是一个一个开始，然后把所有点走一遍，再判断是否成立，这个是可以的，也是最正常的思维，但复杂度O(n2)
这就是第一种解法
不得不优化。
有如下规律，如果一串数字，总和大于等于0，则必存在某一点，从这点开始，一真累加，每次都是非负数
这样，我们只需要判断最后的结果是否为非负就知道是否存在解，即total>=0
要找到开始的那个点，就判断是否是一直>=0，