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.分析
这道题使用暴力搜索的办法应该会超时，需要寻找一种比较省时的方法。首先说一个结论，如果gas的总和小于cost的总和，那么必定没有解，返回-1即可。如果gas的总和大于cos的总和，那么必定有解。
首先计算diff数组得到diff[i] = gas[i] - cost[i]，表示这一段路净剩多少燃油。start起点的位置必定是跑一圈之后净剩的燃油始终为正数，否则中途如果出现负数就会失败。所以我们要找一个位置使得以其未开始净剩的燃油不小于0。
我们从0开始遍历，如果计算当前净剩的燃油数为负，那么说明0到该位置都不可能当做起点，那么就选择下一个当做起点，同时净剩燃油数置0。如果燃油数始终为正，那么就继续遍历，直到遍历整个数组后得到的start位置就是最终的位置。
gas12345cost34512diff-2-2-233sum-2 (置0） -2(置0) -2（置0） 3 6
Code
class Solution { public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int len = gas.size(); vector<int> diff(len, 0); int sum = 0; for (int i = 0; i < len; i ++) { diff[i] = gas[i] - cost[i]; sum += diff[i]; } if (sum < 0) return -1; int start = 0; sum = 0; for (int i = 0; i < len; i ++) { sum += diff[i]; if (sum < 0) { start = i + 1; sum = 0; } } return start; } };运行效率
Runtime: 8 ms, faster than 95.66% of C++ online submissions for Gas Station.
Memory Usage: 9.3 MB, less than 5.50% of C++ online submissions for Gas Station.
