题目描述
环形路上有n个加油站，第i个加油站的汽油量是gas[i].
你有一辆车，车的油箱可以无限装汽油。从加油站i走到下一个加油站（i+1）花费的油量是cost[i]，你从一个加油站出发，刚开始的时候油箱里面没有汽油。
求从哪个加油站出发可以在环形路上走一圈。返回加油站的下标，如果没有答案的话返回-1。
注意：
答案保证唯一。
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.
输入
[2,3,1],[3,1,2]
输出
1
class Solution {
public:
/**
*
* @param gas int整型vector
* @param cost int整型vector
* @return int整型
*/
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
// write code here
int start=gas.size()-1;
int end=0;
int sum=gas[start]-cost[start];
while (start>end)
{
if (sum>=0){
sum+=gas[end]-cost[end];
++end;
}else {
--start;
sum+=gas[start]-cost[start];
}
}
return sum>=0 ?start:-1;
}
};