摘要:
用竞赛的方法,只要n+logn-2次比较Winner competes against ⌈lg n⌉ opponents.Clearly, the tournament performs n − 1 comparisons————⊲ Each comparison removes an element FindMax on line 20 does ⌈lg n⌉ − 1 comparisons————⊲ Winner’s list has ⌈lg n⌉ elements Total comparison count: n + ⌈lg n⌉ − 2 阅读全文
posted @ 2011-05-24 21:34
法兰天蝎
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该文被密码保护。 阅读全文
posted @ 2011-05-24 02:47
法兰天蝎
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