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.

 

solution:

如果从i出发可以环绕一圈的话，则对于路径上任何一点j，到达j时tank均大于等于零。从index=0为start开始寻找起点，需满足tank += gas[0]-cost[0]>0,如果到达i时，tank += gas[i]-cost[i]<0,则说明从0到i均不能作为起点（i点中gas[i]-cost[i]<0,显然不能做起点），而从0到i-1中，以0为起点时到达i后tank最大，此时仍不满足tank += gas[i]-cost[i]>=0,因此0到i-1均不能作为起点，从i+1开始继续搜索起点。

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