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.
思路: 设置以当前的 sum 和一个 total 的 sum , 如果当前的sum < 0 ,则从这个开始重新开始。 注意 最后检查是否可以到达一圈, 也就是 total >= 0 .
易错点: 最后返回 index + 1, 初始化为index = -1;
public class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int sum = 0, total = 0, index = -1;
for(int i = 0; i < gas.length; i++){
sum += gas[i] - cost[i];
total += gas[i] - cost[i];
if(sum < 0){
sum = 0;
index = i;
}
}
return total >= 0 ? index + 1 : -1;
}
}