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Vivelapaix
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Jun 4, 2020
linked-list.md
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| if not headA or not headB: | ||
| return None | ||
| curA, curB = headA, headB | ||
| flagA, flagB = False, False |
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Ты не прочувствовал решение. Объясни мне, почему эти две переменные на самом деле не нужны
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@Vivelapaix ну да, в принципе можно и с двумя указателями сделать...
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@Vivelapaix Вот решение с двумя указателями. |
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Added LEETCODE "intersection-of-two-linked-lists" solution.