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$O(N^2)$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

Python

// 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;
    }
};

Java

// Gas Station
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int total = 0;
        int j = -1;
        for (int i = 0, sum = 0; i < gas.size(); ++i) {
            sum += gas[i] - cost[i];
            total += gas[i] - cost[i];
            if (sum < 0) {
                j = i;
                sum = 0;
            }
        }
        return total >= 0 ? j + 1 : -1;
    }
};

C++

# Gas Station
# Time complexity O(n), space complexity O(1)
class Solution:
    def canCompleteCircuit(self, gas: list[int], cost: list[int]) -> int:
        total = 0
        j = -1
        sum = 0
        for i in range(len(gas)):
            sum += gas[i] - cost[i]
            total += gas[i] - cost[i]
            if sum < 0:
                j = i
                sum = 0
        return j + 1 if total >= 0 else -1