題目：
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.
題解：
網上搜到的解法。
據說是Bloomberg的面試題。
用兩個變數存+= gas[i] - cost[i]。一個幫助判斷當前這個點作為gas station的起點合不合適，一個幫助判斷總的需求是不是大於供給。如果總的需求大於供給那麼肯定是無解的，如果需求小於等於供給，就可以返回剛才找到的起始點。
程式碼如下：
1 public int canCompleteCircuit(int[] gas, int[] cost) {
2 if (gas==null|| cost==null||gas.length==0||cost.length==0||gas.length!=cost.length)
3 return -1;
4
5 int sum = 0;
6 int total = 0;
7 int index = 0;
8 for(int i = 0; i < gas.length; i++){
9 sum += gas[i]-cost[i];
10 total += gas[i]-cost[i];
11 if(sum < 0){
12 index=i+1;
13 sum = 0;
14 }
15 }
16 if(total<0)
17 return -1;
18 else
19 return index;
20 }
2 if (gas==null|| cost==null||gas.length==0||cost.length==0||gas.length!=cost.length)
3 return -1;
4
5 int sum = 0;
6 int total = 0;
7 int index = 0;
8 for(int i = 0; i < gas.length; i++){
9 sum += gas[i]-cost[i];
10 total += gas[i]-cost[i];
11 if(sum < 0){
12 index=i+1;
13 sum = 0;
14 }
15 }
16 if(total<0)
17 return -1;
18 else
19 return index;
20 }
Reference：http://blog.csdn.net/lbyxiafei/article/details/12183461