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.

//首先，这个题目要明确如果gas[0] + gas[1] + ... + gas[n] >= cost[0] + cost[1] + .. + cost[n]，那么这个题目一定有解。因为，根据条件移项可得：//(gas[0] - cost[0]) + (gas[1] - cost[1]) + ... + (gas[n] - cost[n]) >= 0，由于最终结果大于等于零，因此一定可以通过加法交换律，得到一个序列，使得中间结果都为非负public class Solution {    public int canCompleteCircuit(int[] gas, int[] cost) {        int start = 0;        int currGasRemain = 0;        int totalGasRemain = 0;                for (int curr = 0; curr < gas.length; curr++) {            currGasRemain += gas[curr] - cost[curr];            totalGasRemain += gas[curr] - cost[curr];                        if (currGasRemain < 0) {                start = curr + 1; // 更新到下一站， 因为cost[curr]表示从curr 站到curr+1 站的花销                currGasRemain = 0;            }        }                return totalGasRemain >= 0 ? start : -1;    }}

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