<leetcode> 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.

思路来源于水中的鱼

新建一个油耗数组diff[], 赋值为diff[i]=gas[i]-cost[i] 若存在解，则证明diff[]的和为正值（总补给大于总消耗），若diff[]和为负值，证明无解(消耗大于补给)

设解为k（唯一），则diff[]中从k到diff.length-1的和为正值（可以从k走到最后），并且对于其中任意一个点n，k到n的和均为正值（否则k无法到n）

设一个sum＝sum＋diff[i],当sum小于0的时候从当前位置从新开始计算（sum=0,start=i+1）直到找到一个k使k到(diff.size-1)的和为正值。如果没有返回－1；

找到k后同时需要确定是否有解（即k能否从length-1走到k(loop)）,若diff[]和为正则证明有解，k为唯一解，若为负证明无解返回-1

public class Solution {    public int canCompleteCircuit(int[] gas, int[] cost) {        int[] diff=new int[gas.length];        for(int i=0;i<gas.length;i++){            diff[i]=gas[i]-cost[i];        }        int leftGas=0;        int sum=0;        int start=0;        for(int i=0;i<gas.length;i++){            leftGas+=diff[i];            sum+=diff[i];            if(sum<0){                sum=0;                start=i+1;            }        }        if(start>=gas.length){            return -1;        }        else if(leftGas<0){            return -1;        }        else return start;    }}

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