134. Gas Station

问题描述

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.

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.

题目链接：

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思路分析

给两个数组，一个是加油站能加的油，另一个是到下一个加油站需要的油，问从那个加油站开始能绕所有加油站一周。

用两个循环遍历判断。如果能回来就是可以的，否则不行返回-1。

代码

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int n = gas.size();
        for(int i = 0; i < n; i++){
            if (gas[i] < cost[i]) continue;
            int start = i, j = i, tank = gas[i];
            do{
                tank -= cost[j];
                if(tank < 0)
                    break;
                j = (j+1) % n;
                tank += gas[j];
            }while(j != start);
            if(j == start) return start;
        }
        return -1;
    }
};

时间复杂度：$O(n^2)$
空间复杂度：$O(1)$

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反思

然而，事情并没有那么复杂，可以在线性时间完成。因为如果cost加起来大于gas，那肯定无法完成；如果可以完成一圈，那么假设从最后一个位置开始，向前遍历，end从0开始向后遍历，必然能够找到一个使得sum>0的位置，就是我们应该开始地方。

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int start = gas.size() - 1;
        int end = 0;
        int sum = gas[start] - cost[start];
        while (start > end){
            if (sum >= 0){
                sum += gas[end] - cost[end];
                end++;
            }
            else{
                --start;
                sum += gas[start] - cost[start];
            }
        }
        return sum >= 0 ? start : -1;
    }
};