leetcode-problemset/leetcode-cn/problem (Chinese)/两个字符串的切换距离 [shift-distance-between-two-strings].html

<p>给你两个长度相同的字符串&nbsp;<code>s</code> 和&nbsp;<code>t</code>&nbsp;，以及两个整数数组&nbsp;<code>nextCost</code> 和&nbsp;<code>previousCost</code>&nbsp;。</p>

<p>一次操作中，你可以选择 <code>s</code>&nbsp;中的一个下标 <code>i</code>&nbsp;，执行以下操作 <strong>之一</strong>&nbsp;：</p>

<ul>
	<li>将&nbsp;<code>s[i]</code>&nbsp;切换为字母表中的下一个字母，如果&nbsp;<code>s[i] == 'z'</code>&nbsp;，切换后得到&nbsp;<code>'a'</code>&nbsp;。操作的代价为&nbsp;<code>nextCost[j]</code>&nbsp;，其中&nbsp;<code>j</code>&nbsp;表示&nbsp;<code>s[i]</code>&nbsp;在字母表中的下标。</li>
	<li>将&nbsp;<code>s[i]</code>&nbsp;切换为字母表中的上一个字母，如果&nbsp;<code>s[i] == 'a'</code>&nbsp;，切换后得到&nbsp;<code>'z'</code>&nbsp;。操作的代价为&nbsp;<code>previousCost[j]</code>&nbsp;，其中&nbsp;<code>j</code> 是&nbsp;<code>s[i]</code>&nbsp;在字母表中的下标。</li>
</ul>

<p><strong>切换距离</strong>&nbsp;指的是将字符串 <code>s</code>&nbsp;变为字符串 <code>t</code>&nbsp;的 <strong>最少</strong>&nbsp;操作代价总和。</p>

<p>请你返回从 <code>s</code>&nbsp;到 <code>t</code>&nbsp;的 <strong>切换距离</strong>&nbsp;。</p>

<p>&nbsp;</p>

<p><strong class="example">示例 1：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b>s = "abab", t = "baba", nextCost = [100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], previousCost = [1,100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]</span></p>

<p><span class="example-io"><b>输出：</b>2</span></p>

<p><b>解释：</b></p>

<ul>
	<li>选择下标&nbsp;<code>i = 0</code>&nbsp;并将&nbsp;<code>s[0]</code>&nbsp;向前切换 25 次，总代价为 1 。</li>
	<li>选择下标&nbsp;<code>i = 1</code>&nbsp;并将&nbsp;<code>s[1]</code>&nbsp;向后切换 25 次，总代价为 0 。</li>
	<li>选择下标&nbsp;<code>i = 2</code>&nbsp;并将&nbsp;<code>s[2]</code>&nbsp;向前切换 25 次，总代价为 1 。</li>
	<li>选择下标&nbsp;<code>i = 3</code>&nbsp;并将&nbsp;<code>s[3]</code>&nbsp;向后切换 25 次，总代价为 0 。</li>
</ul>
</div>

<p><strong class="example">示例 2：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b>s = "leet", t = "code", nextCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], previousCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]</span></p>

<p><span class="example-io"><b>输出：</b>31</span></p>

<p><b>解释：</b></p>

<ul>
	<li>选择下标&nbsp;<code>i = 0</code>&nbsp;并将&nbsp;<code>s[0]</code>&nbsp;向前切换 9 次，总代价为 9 。</li>
	<li>选择下标&nbsp;<code>i = 1</code>&nbsp;并将&nbsp;<code>s[1]</code>&nbsp;向后切换 10 次，总代价为 10 。</li>
	<li>选择下标&nbsp;<code>i = 2</code> 并将&nbsp;<code>s[2]</code>&nbsp;向前切换 1 次，总代价为 1 。</li>
	<li>选择下标 <code>i = 3</code> 并将&nbsp;<code>s[3]</code>&nbsp;向后切换 11 次，总代价为 11 。</li>
</ul>
</div>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>1 &lt;= s.length == t.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> 和&nbsp;<code>t</code>&nbsp;都只包含小写英文字母。</li>
	<li><code>nextCost.length == previousCost.length == 26</code></li>
	<li><code>0 &lt;= nextCost[i], previousCost[i] &lt;= 10<sup>9</sup></code></li>
</ul>