[leetcode]gas-station加油站（貪心）

問題描述：

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.

解決思路：

將start放在最後一個加油站，end放在第一個。從start出發，如果油量可以走到下一站，就向後走（end++，代表耗油），如果油量不夠了，start要向後退（start--，代表去加油）。如果start和end重合的時候還有油（>=0），題目又規定了一定是唯一解，這就說明當前start位置就是要求的位置，如果油量不夠（<0）則無解，返回-1。

具體實現：

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;
    }
};

參考了大神們的做法，真的巧妙！繼續加油鴨，編程能力會變好的！會變強的！