【Grind75Hard】8問目 23. Merge k Sorted Lists #67
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oda
reviewed
May 24, 2024
| for i, node in enumerate(lists): | ||
| if not node: | ||
| continue | ||
| heapq.heappush(min_heap, (node.val, i, node)) |
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これがないと、定義されていないListNodeの比較になって問題が出るのですかね。
Owner
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問題文上は不要です。
これがないと、定義されていないListNodeの比較になって問題が出るのですかね。
おっしゃる通りで、同値の比較のために入れていました。
この旨はコメントで書いておかないと、疑問を持たれてしまいますね。
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同じ値があったときに、オブジェクトの < が定義されていないからエラーということですね。
liquo-rice
reviewed
May 24, 2024
| for i, node in enumerate(lists): | ||
| if not node: | ||
| continue | ||
| heapq.heappush(min_heap, (node.val, i, node)) |
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これがないと、定義されていないListNodeの比較になって問題が出るのですかね。
| node = sentinel | ||
| while min_heap: | ||
| val, index, smallest_head = heapq.heappop(min_heap) | ||
| node.next = ListNode(val) |
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こういうことですかね。
val, index, smallest_head = heapq.heappop(min_heap)
node.next = smallest_head
node = node.next既存のノードのnextは必ず上書きされるので、新しく作った方が分かりやすいかなと考えてました。
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問題
https://leetcode.com/problems/merge-k-sorted-lists/