Gas Station (M)

题目

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.

题意

有n个汽油站，第i个汽油站存有油gas[i]，从第i个汽油站前往下一个汽油站的油耗为cost[i]，求一辆车能否从某一个汽油站开始(汽车初始油量为0)走遍全部n个汽油站。

思路

参见 [leetcode] 134. Gas Station 解题报告。

代码实现

Java

class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int start = 0, left = 0, debt = 0;
        for (int i = 0; i < gas.length; i++) {
            left += gas[i] - cost[i];
            if (left < 0) {
                debt += left;
                start = i + 1;
                left = 0;
            }
        }
        return left + debt >= 0 ? start : -1;
    }
}

JavaScript

/**
 * @param {number[]} gas
 * @param {number[]} cost
 * @return {number}
 */
var canCompleteCircuit = function(gas, cost) {
    let left = []
    let remain = 0, owe = 0, start = 0
    gas.forEach((item, i) => left.push(item - cost[i]))
    for (let i = 0; i < gas.length;i++) {
        remain += left[i]
        if (remain < 0) {
            start = i + 1
            owe += remain
            remain = 0
        }
    }
    if (remain + owe >= 0) {
        return start
    }
    return -1
}

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