public class Solution {public int canCompleteCircuit(int[] gas, int[] cost) {int sum=0, total=0;int j=-1;for(int i=0; i<gas.length; i++){sum += gas[i]-cost[i];//单个节点剩余的汽油，判断是否是有效节点total += gas[i]-cost[i];//总共剩余的汽油if(sum < 0){j=i;sum=0;} }return total >=0 ? j+1:-1;}
}分析：求唯一的一个有效解：从任意一个点出发，如果不能走回来，那么这个解无效；从下一个点出发，继续验证。直到找到唯一有效解。这个对每个点进行模拟的复杂度是O（N^2）. 因为题目说如果有解，就是唯一解。这里有一个投机取巧的方法：总共消耗的汽油不超过总共提供的汽油，并且记录的有效节点就是那个唯一解。
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.