leetcode 134. gas station

Topic link
Ideas ： One traverse
analysis ： Then start from each site and try
however , Start site , The fuel that can be added must be greater than or equal to the fuel consumption to the next station .
namely gas[i]>=cost[i]
And because the answer is only , So the amount of oil you can add is definitely not 0
namely gas[i]!=0
Assuming that the x Leaves at , It's the farthest y, Suppose you don't go through all the sites .
So the next starting point , No x Post site , It should be y The following sites .
because x and y The middle site , It's past , I've judged that I can't .
Code ：

class Solution {
    
    public int canCompleteCircuit(int[] gas, int[] cost) {
    
        int len = gas.length;
        int max =0;
        int start =0;

        for(int i=0;i<len;i++){
    
            // The starting point must be   The oil you can add is more than the gasoline you spend   And because the answer is only , So the starting point is definitely not refueling 0
            if(gas[i]>=cost[i] && gas[i]!=0){
    
                //start Record the starting point 
                start = i;
                // Record the amount of oil at this time , It must have been 0
                int t =0;

                // Record the sites you passed 
                int count=0;

                // Judge whether there is enough oil at the next station   also   Judge whether you have run through all sites 
                while(t+gas[start%len]-cost[start%len]>=0 && count<=len){
    
                    t = t+gas[start%len]-cost[start%len];
                    start++;
                    count++;
                }
                // If you run through all the sites , Then you'll find the answer 
                if(count>=len){
    
                    return i;
                }else{
    
                    // Otherwise, start from the next service station that you can't get to , Because starting from the service station in front , I've tried , Can't go through all sites , Think about it 
                    i = i+count;
                }
            }
        }
        return -1;
    }
}