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.
沿着环形路线有N个加油站，其中站i的气体量是气体[i]。
你有一辆带有无限油箱的汽车，从车站i到下一站（i + 1）需要花费成本[i]。 您可以在其中一个加油站开始使用空罐。
如果您可以顺时针方向绕电路一次返回起始加油站的索引，否则返回-1。
注意：
如果存在解决方案，则保证是唯一的。两个输入数组都是非空的并且具有相同的长度。输入数组中的每个元素都是非负整数。 2，题目思路对于这道题，判断汽车是否能走完环形加油站。
原本以为是一个路径可行的规划问题，后来发现整个路径是一个环路而且必须是相邻两两才有可能走的。
因此，在这个问题的解决上：
首先：如果总的代价要比总的汽油量小，说明一定存在着解。同时，因为有且只要一个解决方案，因此当我们找到第一个i->i+1可走的方法时（即方案的开始位置），就一定是最终的解决办法。
也就是说，不存在i->i+1可行，突然i+2->i+3又不可行了。
另外，需要注意的是，这个可走性和起始节点有很大关系：
并不是说3->4->5->0->1->2->3可走，2->3->4->5->0->1就也可走。事实上，如果我们从i=0开始搜索，0->1肯定不可走，1->2也肯定不可走。
其次，tank（油箱）的值一定不能为负值，即如果从A->B，如果tank值变为负，说明从A->B一定不可达，这时，让start设为B，并清空tank。
3，代码实现 static auto speedup = [](){ ios::sync_with_stdio(false); cin.tie(nullptr); return nullptr;}();class Solution {public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int sumGas = 0; int sumCost = 0; int start = 0; int tank = 0; for(int i = 0;i<gas.size();i++){ sumGas += gas[i]; sumCost += cost[i]; tank += gas[i] - cost[i]; if(tank < 0){ //此时不可到当前port，意义下一个位置作为开始位置 start = i+1; tank = 0; } } //如果总的汽油小于总代价 if(sumGas < sumCost) return -1; else return start; }};