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leetcode-problemset/leetcode-cn/problem (Chinese)/统计只差一个字符的子串数目 [count-substrings-that-differ-by-one-character].html
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<p>给你两个字符串 <code>s</code> 和 <code>t</code> ,请你找出 <code>s</code> 中的非空子串的数目,这些子串满足替换 <strong>一个不同字符</strong> 以后,是 <code>t</code> 串的子串。换言之,请你找到 <code>s</code> 和 <code>t</code> 串中 <strong>恰好</strong> 只有一个字符不同的子字符串对的数目。</p>
<p>比方说, <code>"<strong>compute</strong>r"</code> 和 <code>"<strong>computa</strong>tion"</code> 加粗部分只有一个字符不同: <code>'e'</code>/<code>'a'</code> ,所以这一对子字符串会给答案加 1 。</p>
<p>请你返回满足上述条件的不同子字符串对数目。</p>
<p>一个 <strong>子字符串</strong> 是一个字符串中连续的字符。</p>
<p> </p>
<p><strong>示例 1</strong></p>
<pre>
<b>输入:</b>s = "aba", t = "baba"
<b>输出:</b>6
<strong>解释:</strong>以下为只相差 1 个字符的 s 和 t 串的子字符串对:
("<strong>a</strong>ba", "<strong>b</strong>aba")
("<strong>a</strong>ba", "ba<strong>b</strong>a")
("ab<strong>a</strong>", "<strong>b</strong>aba")
("ab<strong>a</strong>", "ba<strong>b</strong>a")
("a<strong>b</strong>a", "b<strong>a</strong>ba")
("a<strong>b</strong>a", "bab<strong>a</strong>")
加粗部分分别表示 s 和 t 串选出来的子字符串。
</pre>
<strong>示例 2</strong>
<pre>
<b>输入:</b>s = "ab", t = "bb"
<b>输出:</b>3
<strong>解释:</strong>以下为只相差 1 个字符的 s 和 t 串的子字符串对:
("<strong>a</strong>b", "<strong>b</strong>b")
("<strong>a</strong>b", "b<strong>b</strong>")
("<strong>ab</strong>", "<strong>bb</strong>")
加粗部分分别表示 s 和 t 串选出来的子字符串。
</pre>
<strong>示例 3</strong>
<pre>
<b>输入:</b>s = "a", t = "a"
<b>输出:</b>0
</pre>
<p><strong>示例 4</strong></p>
<pre>
<b>输入:</b>s = "abe", t = "bbc"
<b>输出:</b>10
</pre>
<p> </p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 <= s.length, t.length <= 100</code></li>
<li><code>s</code> 和 <code>t</code> 都只包含小写英文字母。</li>
</ul>