134. Gas Station
134. Gas Station
题目
There are n
gas stations along a circular route, where the amount of gas at the ith
station is gas[i]
.
You have a car with an unlimited gas tank and it costs cost[i]
of gas to travel from the ith
station to its next (i + 1)th
station. You begin the journey with an empty tank at one of the gas stations.
Given two integer arrays gas
and cost
, return the starting gas station 's index if you can travel around the circuit once in the clockwise direction, otherwise return -1
. If there exists a solution, it is guaranteed to be unique
Example 1:
Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
Output: 3
Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.
Example 2:
Input: gas = [2,3,4], cost = [3,4,3]
Output: -1
Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.
Constraints:
n == gas.length == cost.length
1 <= n <= 10^5
0 <= gas[i], cost[i] <= 10^4
题目大意
在一条环路上有 n
个加油站,其中第 i
个加油站有汽油 gas[i]
升。
你有一辆油箱容量无限的的汽车,从第 i
个加油站开往第 i+1
个加油站需要消耗汽油 cost[i]
升。你从其中的一个加油站出发,开始时油箱为空。
给定两个整数数组 gas
和 cost
,如果你可以按顺序绕环路行驶一周,则返回出发时加油站的编号,否则返回 -1
。如果存在解,则 保证 它是 唯一 的。
解题思路
这个问题可以通过一次遍历来解决,遍历过程中记录总加油量与总消耗量的差值,同时维护一个最小差值的变量和一个起始索引。
- 初始化变量
sum
表示总加油量与总消耗量的差值,变量minSum
表示最小的差值,变量start
表示起始索引,以及变量n
表示加油站的数量。 - 遍历每个加油站,计算总加油量与总消耗量的差值,并判断是否小于当前最小差值。如果是,则更新最小差值和起始索引。
- 遍历结束后,如果总差值小于
0
,说明无法绕一周,返回-1
;否则,返回起始索引。
- 时间复杂度:
O(n)
,其中n
是加油站的数量。一次遍历所有加油站。 - 空间复杂度:
O(1)
,只使用了常数个额外变量。
代码
/**
* @param {number[]} gas
* @param {number[]} cost
* @return {number}
*/
var canCompleteCircuit = function (gas, cost) {
let sum = 0,
minSum = 0,
start = 0,
n = gas.length;
for (let i = 0; i < n; i++) {
sum += gas[i] - cost[i];
if (sum < minSum) {
minSum = sum;
start = i + 1;
}
}
if (sum < 0) {
return -1;
}
return start == n ? 0 : start;
};
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