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.
此题一看就有思路，不过是n^2的解法，就是轮训每个气站，直到找到符合条件的气站，思路简单但是又果然超时了，代码如下： class Solution { public: bool judge(vector<int>& gas, vector<int>& cost, int k) { if(gas[k] < cost[k]) return false; int curgas = 0; for(int i=k; i<gas.size(); ++i) { curgas += gas[i]; curgas -= cost[i]; //cout<<curgas<<' '; if(curgas < 0) return false; } for(int i=0; i<k; ++i) { curgas += gas[i]; curgas -= cost[i]; //cout<<curgas<<' '; if(curgas < 0) return false; } return true; } int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { if(gas.size() != cost.size()) return -1; if(gas.empty()) return -1; for(int i=0; i<gas.size(); ++i) { if(judge(gas, cost, i)) return i; } return -1; } }; 实在想不到好多解法，就参考别人思路得到ac代码。注意题目中给出的条件：如果有解就确定有一个解，或者没有解。此条件是最重要的，就是说除非所有气站“气量和”小于所有路径消耗“气量和”而没有解外，其他情况下均有一个解。那么我们就可以用排除法来做了。
另外一点很重要：从0到i刚好不可到达，那么从0～i-1都不可到达i。
得到一下代码：
class Solution { public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { if(gas.size() != cost.size()) return -1; if(gas.empty()) return -1; int sum = 0, total = 0, j = 0; for(int i=0; i<gas.size(); ++i) { sum += gas[i] - cost[i]; total += gas[i] - cost[i]; if(sum < 0)//从0到i刚好不可到达，那么从0～i-1都不可到达i { sum = 0; j = i+1; } } if(total < 0) return -1; return j; } };
