134. Gas Station

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 in the clockwise direction, otherwise return -1.

Note:

• If there exists a solution, it is guaranteed to be unique.
• Both input arrays are non-empty and have the same length.
• Each element in the input arrays is a non-negative integer.

Example 1:

Input: 
gas  = [1,2,3,4,5]
cost = [3,4,5,1,2]

Output: 3

Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.

Example 2:

Input: 
gas  = [2,3,4]
cost = [3,4,3]

Output: -1

Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.

 

解题思路：

说有N个加油站，他们是成环状分布，然后给你一辆车，并且告诉你这辆车从这个加油站i ,到下一个加油站 i+1 所要花费的油，也告诉你这个加油站可以给你提供的油。问你能否有这样一个起点，你空箱加油后可以绕一圈回到这个加油站。

一开始我也没有什么好的想法然后我就以每个点为起点，开始看可不可行（嗯，最笨的办法）

然后我run code的时候就超时了：）

看了大佬的解法，并且体会了一下。

油就那么多，如果所有油加起来不到总消耗，那么从哪里走都走不过去。

用total来判断总供给能否满足总消耗

供油和耗油就两种关系：满足也就是供油>=耗油， 不满足: 供油<= 耗油

如果以某点为起点，到另一点时，此时供油-耗油为负的，那么说名起点和这个点之间的所有点都不能做起点

因为假设起点给油过剩，那中途的点都给耗没了，所以我们可以直接将起始点设置为下一个点。

 

要冷静下来多分析一下题意啊朋友们

 

代码：

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