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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.
class Solution { public: int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { int num = gas.size(); int remain = 0;//从第一站开始到本站剩余的gas量 int start = 0; for(int i = 0;i<num;) { if(cost[i]>gas[i]) { i++; continue; } else { start = i;//标记开始站 remain = gas[i]-cost[i]; i++; while(i<num &&gas[i]+remain >= cost[i]) { remain = (gas[i]+remain - cost[i]); i++; } if(i == num) { i = 0; } else { start = i+1; i = start; continue; } while(gas[i]+remain >= cost[i] && i<start) { remain = (gas[i]+remain - cost[i]); i++; } if(i == start) { return start; } else { return -1; } } }//end for return -1; } };
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原文地址：http://www.cnblogs.com/Xylophone/p/3849541.html
