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[Leetcode]134. Gas Station

题目描述

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 in the clockwise direction, otherwise return -1.

Note:

If there exists a solution, it is guaranteed to be unique.
Both input arrays are non-empty and have the same length.
Each element in the input arrays is a non-negative integer.

例子

例子 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.

例子 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.

解题思路

首先这道题说明了只要有解的情况解肯定唯一。所以在有解的情况下,我们思路可以是遍历一遍数字,同时维护一个值表示当前剩的油量,每到一个位置利用 gas[i] - cost[i] 更新油量,当油量小于 0 时,在这之前的位置都不能作为解。因此设油量为 0 在下一个位置重新计算;最后还有通过判断总油量和总花费来判断解是否存在,代码如下:

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class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {

int num_stations = gas.size();
int current_gas = 0, index = -1;
int total_gas = 0, total_cost = 0;
for (int i = 0; i < num_stations; i++) {
total_gas += gas[i];
total_cost += cost[i];

// 当前存油量为 0 且该位置油量大于花费时,可以作为起始点
if (gas[i] >= cost[i] && current_gas == 0) {
index = i;
}

current_gas += gas[i] - cost[i];
// 当前村油量到不了下个位置,以前的起始点作废,下个位置重新计算
if (current_gas < 0) {
index = -1;
current_gas = 0;
}

}

if (total_gas < total_cost) return -1;
return index;

}
};

时间复杂度: O(n)
空间复杂度: O(1)