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Algorithms

There are <code>n</code> gas stations along a circular route, where the amount of gas at the <code>ith</code> station is <code>gas[i]</code>.

You have a car with an unlimited gas tank and it costs <code>cost[i]</code> of gas to travel from the <code>ith</code> station to its next <code>(i + 1)th</code> station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays <code>gas</code> and <code>cost</code>, return the starting gas station&#39;s index if you can travel around the circuit once in the clockwise direction, otherwise return <code>-1</code>. If there exists a solution, it is guaranteed to be unique.

&nbsp;
Example 1:

<pre>
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.
</pre>

Example 2:

<pre>
Input: gas = [2,3,4], cost = [3,4,3]
Output: -1
Explanation:
You can&#39;t start at station 0 or 1, as there is not enough gas to travel to the next station.
Let&#39;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&#39;t travel around the circuit once no matter where you start.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>n == gas.length == cost.length</code></li>
	<li><code>1 &lt;= n &lt;= 105</code></li>
	<li><code>0 &lt;= gas[i], cost[i] &lt;= 104</code></li>
</ul>

canCompleteCircuit

Gas Station

Maximize the Topmost Element After K Moves

maximize-the-topmost-element-after-k-moves

K 次操作后最大化顶端元素

Greedy

Array

在一条环路上有 <code>n</code>&nbsp;个加油站，其中第 <code>i</code>&nbsp;个加油站有汽油&nbsp;<code>gas[i]</code>&nbsp;升。

你有一辆油箱容量无限的的汽车，从第 <code>i</code> 个加油站开往第 <code>i+1</code>&nbsp;个加油站需要消耗汽油&nbsp;<code>cost[i]</code>&nbsp;升。你从其中的一个加油站出发，开始时油箱为空。

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 保证 它是 唯一 的。

&nbsp;

示例&nbsp;1:

<pre>
输入: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
输出: 3
解释:
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>

示例 2:

<pre>
输入: gas = [2,3,4], cost = [3,4,3]
输出: -1
解释:
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>

&nbsp;

提示:

<ul>
	<li><code>gas.length == n</code></li>
	<li><code>cost.length == n</code></li>
	<li><code>1 &lt;= n &lt;= 105</code></li>
	<li><code>0 &lt;= gas[i], cost[i] &lt;= 104</code></li>
</ul>