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Aug 5, 2024
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feat: add solutions to lc problem: No.3244 #3364

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feat: add solutions to lc problem: No.3244
yanglbme Aug 5, 2024
de745a2
fix: update
yanglbme Aug 5, 2024
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fix: update
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@yanglbme
yanglbme committed Aug 5, 2024
commit de745a258fadb9e47a3b6914dedd710eaab2c68c
2 changes: 1 addition & 1 deletion solution/3200-3299/3244.Shortest Distance After Road Addition Queries II/README.md
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Expand Up @@ -88,7 +88,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3244.Sh

### 方法一：贪心 + 记录跳转位置

我们定义一个长度为 $n - 1$ 的数组 $\textit{nxt}$，其中 $\textit{nxt}[i]$ 表示从城市 $i$ 可以到达的下一个城市的编号。初始时，$\textit{nxt}[i] = i + 1$。
我们定义一个长度为 $n - 1$ 的数组 $\textit{nxt}$，其中 $\textit{nxt}[i]$ 表示从城市 $i$ 可以到达的下一个城市的编号。初始时 $\textit{nxt}[i] = i + 1$。

对于每次查询 $[u, v]$，如果此前已经连通了 $u'$ 和 $v'$，且 $u' <= u < v <= v'$，那么我们可以跳过这次查询。否则，我们需要将 $nxt[u]$ 到 $nxt[v - 1]$ 这些城市的下一个城市编号设置为 $0$，并将 $nxt[u]$ 设置为 $v$。

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