Week4 134. Gas Station

题目

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.

分析

如果用最简单的方法所有点遍历一次，把每个点都作为起点，不过这样子复杂度会比较高。

所以我们现在考虑只需要遍历一次的方法 
经过每个油站对车来讲就是只有两种情况，即加油或耗油，于是我们构造数组leftGas[i] = add[i] - cost[i].表示经过一个油站对油量的影响.
现在考虑从油站i出发一直到油站p油箱油<0, 
即:sum1>=０ 
sum2 = sum1 + leftGas[p]<０

那么如果我们从i出发到不了p，只需要考虑从p出发就好了,这样只需要遍历一次数组就可以得到答案.

代码

class Solution {public:    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {        int l = gas.size();        if (l <= 0) return -1;        int s = 0;        int curSum = 0;        int b = 0 ,total = 0;        for (int i = 0; i < l; ++i) {            curSum += (gas[i] - cost[i]);            total += (gas[i] - cost[i]);            if (curSum >= 0) {            } else {                b = i+1;                curSum = 0;            }        }        if (total >= 0) return b;        else return -1;    }};