Problem

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.

Example

Given 4 gas stations with gas[i]=[1,1,3,1], and the cost[i]=[2,2,1,1]. The starting gas station's index is 2.

Challenge

O(n) time and O(1) extra space

Note

看到这个题目，怎样可以不把它当成一个环路来分析，以及减少多余的空间呢？

例如gas[i]=[1,1,3,1], cost[i]=[2,2,1,1]，我们引入单次剩余油量res，剩余油量和sum，总剩余油量和total，以及可行起点start四个参数。大体上说，只要total > 0，一定有解。下面来找起点，当sum < 0的时候，一定是上一个加油站的单次剩余油量res为负，且与上一次的剩余油量和sum相加依然为负，说明在上一个加油站出现了消耗大于补给的情况，因此一定不能将它作为起点。所以跳过这个耗油量很大的油站，然后将下一个油站作为起点继续判断即可。

Solution

public class Solution {    public int canCompleteCircuit(int[] gas, int[] cost) {        int start = 0, sum = 0, total = 0;        for (int i = 0; i < gas.length; i++) {            int res = gas[i] - cost[i];            //如果上一次的剩余油量之和sum’加上油站油量gas[i-1]，            //依然少于上次次的消耗油量cost[i-1]            if (sum < 0) {                 //更新出发地点为第i个加油站                start = i;                //更新剩余油量之和为到达第i个加油站补给后的剩余油量                sum = res;            }            //否则，将之前的剩余油量之和sum加上本次消耗及补给后的剩余油量res            else {                sum += res;            }            //累计所有油站的可补给油量和总消耗油量之差（res）            total += res;        }        return total >= 0 ? start : -1;    }}

Optimized:

public class Solution {    public int canCompleteCircuit(int[] gas, int[] cost) {        int start = 0, sum = 0, total = 0;        for (int i = 0; i < gas.length; i++) {            int res = gas[i] - cost[i];            sum += res;            total += res;            if (sum < 0) {                sum = 0;                start = i + 1;            }        }        return total >= 0 ? start : -1;    }}