[leetcode]gas-station加油站（贪心）

问题描述：

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.

解决思路：

将start放在最后一个加油站，end放在第一个。从start出发，如果油量可以走到下一站，就向后走（end++，代表耗油），如果油量不够了，start要向后退（start--，代表去加油）。如果start和end重合的时候还有油（>=0），题目又规定了一定是唯一解，这就说明当前start位置就是要求的位置，如果油量不够（<0）则无解，返回-1。

具体实现：

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int start = gas.size()-1;
        int end = 0;
        int sum = gas[start] - cost[start];
        while(start > end){
            if(sum > 0){
                sum += gas[end];
                sum -= cost[end];
                end++;
            }else{
                start--;
                sum += gas[start];
                sum -= cost[start];
            }
        }
        return (sum >= 0)? start : -1;
    }
};

参考了大神们的做法，真的巧妙！继续加油鸭，编程能力会变好的！会变强的！