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.
1 如果总的gas - cost小于零的话,那么没有解返回-1。(总gas> 总cost 是有解的充分便要条件,要是解这道题的关键)
2 如果前面所有的gas - cost加起来小于零,那么前面所有的点都不能作为出发点(存在gas-cost<0,剩下的必然存在gas-cost>0)。
3 选择最佳出发点后,判断从这点出发能否环行一圈。
class Solution { public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int len = gas.size(); int res = -1; int res_gas = 0; int start_station = 0; int total_res = 0; for(int i =0 ; i < len; i++) { res_gas += (gas[i] - cost[i]) ; total_res += (gas[i] - cost[i]) ; if(res_gas < 0) { start_station = i+1; res_gas = 0; } } if(total_res >= 0) { res = start_station; } return res; } };