Gas Station
描述
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.
分析
首先想到的是的解法,对每个点进行模拟。
O(N)
的解法是,设置两个变量,sum
判断当前的指针的有效性;total
则判断整个数组是否有解,有就返回通过sum
得到的下标,没有则返回-1。
代码
// Gas Station
// 时间复杂度O(n),空间复杂度O(1)
public class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int total = 0;
int j = -1;
for (int i = 0, sum = 0; i < gas.length; ++i) {
sum += gas[i] - cost[i];
total += gas[i] - cost[i];
if (sum < 0) {
j = i;
sum = 0;
}
}
return total >= 0 ? j + 1 : -1;
}
};