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.
可以将两个数组看作一个数组,benifit[i]=gas[i]-cost[i] 即为每一段路净所得的油.
两个原则:1. 如果总benifit为负数,则无论如何都开不完一圈。
2. 如果从一个加油站i出发,开到加油站j所属路段的时候油耗尽,那么从i,j之间的任一个加油站出发都会在j路段或j之前路段耗尽油。
基于以上两个原则,需要一个变量total统计benifit, 需要一个下标start标记起始站,初始为0,需要一个变量tank记录油箱。一旦在过程中某个i处发现tank小于0,那么start标记为i+1,意味着从之前start到i之间的加油站都不能作为起始加油站。
最后,查看total是否小于0.如果是则返回-1,不是则返回start.
代码:
1 public int canCompleteCircuit(int[] gas, int[] cost) { 2 int tank = 0; 3 int start = 0; 4 int total = 0; 5 for(int i=0;i<gas.length;i++) 6 { 7 total += gas[i]-cost[i]; 8 tank += gas[i]-cost[i]; 9 if(tank<0) 10 { 11 start = i+1; 12 tank=0; 13 } 14 } 15 return total>=0?start:-1; 16 }