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.

贪心or动态规划？

从gas[0]开始走，遇到油不够的情况，开始station就后退一步，利用以前算的结果可以只算出第一步的油量。直到最后，算出circle的连接点是否满足。

public class Solution {    public int canCompleteCircuit(int[] gas, int[] cost) {                int n = gas.length;                int start = 0, end = 0, have = 0, cur = 0;                for(int i=0; i
     
      = 0) {                end ++;                cur = end;            }            else {                start --;                if(start < 0) {                    start = n - 1;                }                cur = start;            }        }                have += gas[cur] - cost[cur];                return have>=0?start:-1;    }}