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，