※ 引述《RockLee (Now of all times)》之銘言：
: ※ 引述《Leon (Achilles)》之銘言：
: : Sorry, I can't understand your writing.
: : Also, the link seems to be wrong...
: : Can you describe it, step by step.
: : For example, there are only 3 words, (a,b,c)
: : how are you going to find the window, for the document
: : [b b a c b b b b c b b a] ?
: 雖然 pseudo code 可能比較短,
: 但由於 interview Google 時必需寫 actual code,
: 所以我想還是直接用實際的 Java code 表達我的想法.
: (我假設 window 長度指的是包含的字數, 與每個字的長度無關)
: 以 Leon 大給的例子:
: document: b b a c b b b b c b b a
: index: 0 1 2 3 4 5 6 7 8 9 10 11
: occurence a: 2, 11
: occurence b: 0, 1, 4, 5, 6, 7, 9, 10
: occurence c: 3, 8
: 執行過程及結果會是:
: windowBegin = 0; windowEnd = 3
: windowBegin = 1; windowEnd = 3
: windowBegin = 2; windowEnd = 4
: windowBegin = 3; windowEnd = 11
: windowBegin = 4; windowEnd = 11
: windowBegin = 5; windowEnd = 11
: windowBegin = 6; windowEnd = 11
: windowBegin = 7; windowEnd = 11
: windowBegin = 8; windowEnd = 11
: bestBegin = 1
: bestEnd = 3
:

找出對應當前start的ending是O(logK)，他在移除start換下一個start時，補上目前start的list中下一個index是必要而且保證是合法的window，TreeMap中總是有K個index找ending只是找出max index而已

我看不懂你寫甚麼