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  • 劍指offer/LintCode12_最小棧

    劍指offer/LintCode12_最小棧 聲明 文章均為本人技術(shù)筆記,轉(zhuǎn)載請(qǐng)注明出處https://segmentfault.com/u/yzwall 解題思路 實(shí)現(xiàn)功能: 實(shí)現(xiàn)一個(gè)最小棧,要求push(element),pop(),min()操作均為$O(1)$復(fù)雜度, 解題思路 用棧stack存儲(chǔ)數(shù)據(jù); 用最小...

    Betta 評(píng)論0 收藏0
  • 劍指offer/LintCode40_用兩個(gè)棧模擬隊(duì)列

    劍指offer/LintCode40_用兩個(gè)棧模擬隊(duì)列 聲明 文章均為本人技術(shù)筆記,轉(zhuǎn)載請(qǐng)注明出處https://segmentfault.com/u/yzwall 解題思路 實(shí)現(xiàn)功能: 用兩個(gè)棧模擬實(shí)現(xiàn)一個(gè)隊(duì)列的push(element),pop()和top()操作; 解題思路 假設(shè)有兩個(gè)棧stack1, stack2 ...

    bawn 評(píng)論0 收藏0
  • 劍指offer/LintCode494_用兩個(gè)隊(duì)列實(shí)現(xiàn)一個(gè)棧

    劍指offer/LintCode494_用兩個(gè)隊(duì)列實(shí)現(xiàn)一個(gè)棧 聲明 文章均為本人技術(shù)筆記,轉(zhuǎn)載請(qǐng)注明出處https://segmentfault.com/u/yzwall 解題思路 實(shí)現(xiàn)功能: 用兩個(gè)隊(duì)列實(shí)現(xiàn)一個(gè)棧,實(shí)現(xiàn)push(element),pop(),top()和isEmpty()方法; 解題思路 假設(shè)有隊(duì)...

    rose 評(píng)論0 收藏0
  • 劍指offer/LeetCode146/LintCode134_LRU緩存實(shí)現(xiàn)

    劍指offer/LeetCode146/LintCode134_LRU緩存實(shí)現(xiàn) 聲明 文章均為本人技術(shù)筆記,轉(zhuǎn)載請(qǐng)注明出處[1] https://segmentfault.com/u/yzwall[2] blog.csdn.net/j_dark/ 解題思路 LRU緩存兩種功能: get(key):獲取key的對(duì)應(yīng)value,不存在返回-1 set(key, value)(lintcod...

    you_De 評(píng)論0 收藏0
  • LintCode547/548_求數(shù)組交集不同解法小結(jié)

    LintCode547/548_求數(shù)組交集不同解法小結(jié) [TOC] 聲明 文章均為本人技術(shù)筆記,轉(zhuǎn)載請(qǐng)注明出處:[1] https://segmentfault.com/u/yzwall[2] blog.csdn.net/j_dark/ LintCode547:求數(shù)組交集_要求元素不重復(fù) LintCode547,給出兩個(gè)數(shù)組,求二者交集且元素...

    gxyz 評(píng)論0 收藏0
  • 兩數(shù)之和問(wèn)題各變種多解法小結(jié)

    ...明出處:[1] https://segmentfault.com/u/yzwall[2] blog.csdn.net/j_dark/ LintCode_56:兩數(shù)之和等于target 題目大意:給出未排序數(shù)組nums和指定目標(biāo)target,返回?cái)?shù)組中兩數(shù)之和$= target$的組合元素下標(biāo)[index1, index2], 要求下標(biāo)從1開始,而且$index1 < index...

    lentoo 評(píng)論0 收藏0
  • [LintCode/LeetCode] Word Break

    ...ace-separated sequence of one or more dictionary words. Example Given s = lintcode, dict = [lint, code]. Return true because lintcode can be break as lint code. Note 基礎(chǔ)動(dòng)規(guī)題目,有幾個(gè)細(xì)節(jié)要注意。dp[0]是指,當(dāng)s為空的時(shí)候...

    dunizb 評(píng)論0 收藏0
  • [LeetCode/LintCode] Top K Frequent Words

    ... res.add(0, queue.poll().getKey()); } return res; } } LintCode version Problem Find top k frequent words with map reduce framework. The mappers key is the document id, value is...

    0x584a 評(píng)論0 收藏0
  • [LintCode/LeetCode] Implement Trie

    ...sume that all inputs are consist of lowercase letters a-z. Example insert(lintcode) search(code) // return false startsWith(lint) // return true startsWith(linterror) // return false insert(linterr...

    付永剛 評(píng)論0 收藏0
  • Lintcode Coins in a line

    LintCode: coins in a line I 有 n 個(gè)硬幣排成一條線。兩個(gè)參賽者輪流從右邊依次拿走 1 或 2 個(gè)硬幣,直到?jīng)]有硬幣為止。拿到最后一枚硬幣的人獲勝。 請(qǐng)判定 第一個(gè)玩家 是輸還是贏? n = 1, 返回 true.n = 2, 返回 true.n = 3, 返回 false.n...

    itvincent 評(píng)論0 收藏0
  • [LintCode/LeetCode] First Unique Character in a S

    ... and return its index. If it doesnt exist, return -1. Example Given s = lintcode, return 0. Given s = lovelintcode, return 2. Tags Amazon Microsoft Bloomberg Solution a fast way class Solution { ...

    Xufc 評(píng)論0 收藏0
  • LintCode: Max Tree

    ...l 很類似, 所以第一反應(yīng)使用遞歸做。遞歸的解法過(guò)不了lintcode,會(huì)顯示超時(shí): class Solution: @param: A: Given an integer array with no duplicates. @return: The root of max tree. def maxTree(self, A): d...

    ivan_qhz 評(píng)論0 收藏0
  • LintCode Coins in a line III

    LintCode Coins in a line III There are n coins in a line. Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount ...

    focusj 評(píng)論0 收藏0
  • [LintCode] Implement Trie

    ...lement a trie with insert, search, and startsWith methods. Example insert(lintcode) search(code) // return false startsWith(lint) // return true startsWith(linterror) // return false insert(linterr...

    chengjianhua 評(píng)論0 收藏0
  • Lintcode Coins in a line II

    LintCode Coins in a line II 有 n 個(gè)不同價(jià)值的硬幣排成一條線。兩個(gè)參賽者輪流從左邊依次拿走 1 或 2 個(gè)硬幣,直到?jīng)]有硬幣為止。計(jì)算兩個(gè)人分別拿到的硬幣總價(jià)值,價(jià)值高的人獲勝。 請(qǐng)判定 第一個(gè)玩家 是輸還是贏? 樣例給...

    2shou 評(píng)論0 收藏0

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