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Approach #2: Double Linked List + TreeMap [Accepted] Intuition Using structures like Array or Stack will never let us popMax quickly. We turn our atte 阅读全文
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Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c. Example 1: Input: 5 Output: True Explanation: 1 * 1 + 2 * 2 = 5 Example 2: Input: ... 阅读全文
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array不是sorted的，里面可能有负数。解法也是从两边往中间搞，用两个数组left和right存原数组从左向右和从右向左的元素和。然后按index来比较left和right两个数组里的元素，有相等的就是balance point. 阅读全文
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Write a function to find the longest common prefix string amongst an array of strings.第二遍做法：时间复杂度应该是O(m*n)，m表示字符串的最大长度，n表示字符串的个数，空间复杂度应该是O(m),即字符串的长度 public class Solution { public String longest... 阅读全文
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二分法逼近，然后找个误差值到时候跳出循环 阅读全文
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Given an unsorted integer array, find the first missing positive integer. For example, Given [1,2,0] return 3, and [3,4,-1,1] return 2. Your algorithm 阅读全文
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Given a tree string expression in balanced parenthesis format:[A[B[C][D]][E][F]].Construct a tree and return the root of the tree. A / | \ B E F / \ C 阅读全文
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private static String convertBinary(int sum) { StringBuffer binary = new StringBuffer(); while (true) { binary.insert(0, sum % 2); sum = sum / 2... 阅读全文
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we start from the leftmost point (or point with minimum x coordinate value) and we keep wrapping points in counterclockwise direction. The big questio 阅读全文
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我就想了个递归, 还是没有区分掉一些重复的情况，worst case O(2^n)基本同暴力解 Map> allSubSet = new HashMap(); Set getAllPalidrome(String s, int x, int y){ int ind = x * s.length() + y; if(allSubSet.constainsKey(ind)) return a... 阅读全文
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每次移动3个pointer 里面最小的那个就好了。记录整个过程找最近的3个数。3个数之中median没意义，距离主要由最小和最大决定。基本就是这个思路。很快编完就过了编程阶段。 阅读全文
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Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest l 阅读全文
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code 不难，multi dimension array 的求 summation。 之前准备地里的LinkedIn高频题一个没碰到。。。 /** Suppose you are given a class that implements a k-dimensional array * interface and you want to perform an operation that ... 阅读全文
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public static int LPS(int[] a) { int[][] dp = new int[a.length][a.length]; for (int i = 0; i = 0; i--) { for (int j = i + 1; j < dp.length; j++) { if (a[i] == a[j]) { dp[i][j] = dp[... 阅读全文
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import java.util.*; import java.util.concurrent.locks.Condition; import java.util.concurrent.locks.Lock; import java.util.concurrent.locks.ReentrantLock; public class BlockQueue { public int cap... 阅读全文
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import java.util.*; import java.util.concurrent.locks.*; public class H2O { private int HCount = 0; private int OCount = 0; private Lock lock = new ReentrantLock(); private Condition condH = lo... 阅读全文
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第一道是设计一个有选择性的iterator，类型T。面试官给的API里有一个自定selector，selector里有差不多叫boolean isOK(T t)方法。设计一个iterator每次调用hasNext()，返回接下来是否能取到合格的T对象；每次调用next()，返回下一个合格的T对象。我 阅读全文
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public class MidStack { static class Node{ T val; Node next = null; Node pre = null; public Node(T val){ this.val = val; } } private int size = 0; private Node head = null; privat... 阅读全文
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这道题要求in-place做法，不能使用extra space, 那么，做法跟Rotate Array那道题非常相似 （1）reverse the whole array （2）reverse each subarray seperated by ' ' 注意不要忘了reverse最后一个word 阅读全文
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/* Suppose you have a long flowerbed in which some of the plots are planted and some are not. However, flowers cannot be planted in adjacent plots - they would compete for water and b... 阅读全文