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Description
tab: English
<p>There are <code>n</code> gas stations along a circular route, where the amount of gas at the <code>i<sup>th</sup></code> station is <code>gas[i]</code>.</p>
<p>You have a car with an unlimited gas tank and it costs <code>cost[i]</code> of gas to travel from the <code>i<sup>th</sup></code> station to its next <code>(i + 1)<sup>th</sup></code> station. You begin the journey with an empty tank at one of the gas stations.</p>
<p>Given two integer arrays <code>gas</code> and <code>cost</code>, return <em>the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return</em> <code>-1</code>. If there exists a solution, it is <strong>guaranteed</strong> to be <strong>unique</strong>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> gas = [1,2,3,4,5], cost = [3,4,5,1,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong>
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.
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> gas = [2,3,4], cost = [3,4,3]
<strong>Output:</strong> -1
<strong>Explanation:</strong>
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.
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>n == gas.length == cost.length</code></li>
<li><code>1 <= n <= 10<sup>5</sup></code></li>
<li><code>0 <= gas[i], cost[i] <= 10<sup>4</sup></code></li>
<li>The input is generated such that the answer is unique.</li>
</ul>
---
[submissions](https://leetcode.com/problems/gas-station/submissions/) | [solutions](https://leetcode.com/problems/gas-station/solutions/)
tab: 中文
<p>在一条环路上有 <code>n</code> 个加油站，其中第 <code>i</code> 个加油站有汽油 <code>gas[i]</code><em> </em>升。</p>
<p>你有一辆油箱容量无限的的汽车，从第<em> </em><code>i</code><em> </em>个加油站开往第<em> </em><code>i+1</code><em> </em>个加油站需要消耗汽油 <code>cost[i]</code><em> </em>升。你从其中的一个加油站出发，开始时油箱为空。</p>
<p>给定两个整数数组 <code>gas</code> 和 <code>cost</code> ，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 <code>-1</code> 。如果存在解，则 <strong>保证</strong> 它是 <strong>唯一</strong> 的。</p>
<p> </p>
<p><strong>示例 1:</strong></p>
<pre>
<strong>输入:</strong> gas = [1,2,3,4,5], cost = [3,4,5,1,2]
<strong>输出:</strong> 3
<strong>解释:
</strong>从 3 号加油站(索引为 3 处)出发，可获得 4 升汽油。此时油箱有 = 0 + 4 = 4 升汽油
开往 4 号加油站，此时油箱有 4 - 1 + 5 = 8 升汽油
开往 0 号加油站，此时油箱有 8 - 2 + 1 = 7 升汽油
开往 1 号加油站，此时油箱有 7 - 3 + 2 = 6 升汽油
开往 2 号加油站，此时油箱有 6 - 4 + 3 = 5 升汽油
开往 3 号加油站，你需要消耗 5 升汽油，正好足够你返回到 3 号加油站。
因此，3 可为起始索引。</pre>
<p><strong>示例 2:</strong></p>
<pre>
<strong>输入:</strong> gas = [2,3,4], cost = [3,4,3]
<strong>输出:</strong> -1
<strong>解释:
</strong>你不能从 0 号或 1 号加油站出发，因为没有足够的汽油可以让你行驶到下一个加油站。
我们从 2 号加油站出发，可以获得 4 升汽油。 此时油箱有 = 0 + 4 = 4 升汽油
开往 0 号加油站，此时油箱有 4 - 3 + 2 = 3 升汽油
开往 1 号加油站，此时油箱有 3 - 3 + 3 = 3 升汽油
你无法返回 2 号加油站，因为返程需要消耗 4 升汽油，但是你的油箱只有 3 升汽油。
因此，无论怎样，你都不可能绕环路行驶一周。</pre>
<p> </p>
<p><strong>提示:</strong></p>
<ul>
<li><code>n == gas.length == cost.length</code></li>
<li><code>1 <= n <= 10<sup>5</sup></code></li>
<li><code>0 <= gas[i], cost[i] <= 10<sup>4</sup></code></li>
<li>输入保证答案唯一。</li>
</ul>
---
[提交记录](https://leetcode.cn/problems/gas-station/submissions/) | [题解](https://leetcode.cn/problems/gas-station/solution/)
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