题目：A robot is located at the top-left corner of a m x n grid (marked ‘Start‘ in the diagram below).The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish‘ in the diagram below).How many possible unique paths are there?Above is a 7 x 3 grid. How many possible unique paths are there?Note: m and n will...
Given two arrays, write a function to compute their intersection. Example: Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2]. Note: Each element in the result should appear as many times as it shows in both arrays. The result can be in any order.题意：求两个数组的交集，每个元素可以出现多次，返回的数组顺序随意。public class Solution { public int[] intersect(int[] nums1, int[] nums2) { ...
这是悦乐书的第292次更新，第310篇原创01 看题和准备今天介绍的是LeetCode算法题中Easy级别的第160题（顺位题号是693）。给定正整数，检查它是否具有交替位：即它的二进制数的任意两个相邻位总是具有不同的值。例如： 输入：5输出：true说明：5的二进制表示是：101 输入：7输出：false说明：7的二进制表示为：111。 输入：11输出：false说明：11的二进制表示是：1011。 输入：10输出：true说明：10的二进制表示是：1010。本次解题使...
【030-Substring with Concatenation of All Words（串联全部单词的子串）】【LeetCode-面试算法经典-Java实现】【全部题目文件夹索引】原题 You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters. For example, given: s: "barfoot...
Given an array of size n, find the majority element. The majority element is the element that appears more than ? n/2 ? times.You may assume that the array is non-empty and the majority element always exist in the array.解题思路：编程之美P130(寻找发帖水王)原题，如果删除两个不同的num，那么剩下的num，最多的元素依旧出现多于? n/2 ? timesJAVA实现如下： public int majorityElement(int[] nums) {LinkedList...
题目描述：Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.Example 1:Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]Return 16The two words can be "abcw", "xtfn".Example 2:Given ["a", "ab", "abc", "d", ...
题目：Given a collection of numbers, return all possible permutations.For example,[1,2,3] have the following permutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1] 思路：最近都是在做DFS有关的，第一时间的反应就是可以用DFS的思想，唯一需要注意的就是在遍历的过程中，要查看是否是已经被选择的数字，这里用Set来进行排查，思路比较简单。 代码：public List<List<Integer>> permute(int[] nums) {Lis...
Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).You may assume that the intervals were initially sorted according to their start times.Example 1:Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].Example 2:Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].This is because the new inter...
题目：给出 n 个数对。 在每一个数对中，第一个数字总是比第二个数字小。现在，我们定义一种跟随关系，当且仅当 b < c 时，数对(c, d) 才可以跟在 (a, b) 后面。我们用这种形式来构造一个数对链。给定一个对数集合，找出能够形成的最长数对链的长度。你不需要用到所有的数对，你可以以任何顺序选择其中的一些数对来构造。示例 :输入: [[1,2], [2,3], [3,4]]输出: 2解释: 最长的数对链是 [1,2] -> [3,4]注意：给出数对的个数在 [1, ...
【LeetCode】Minimum Depth of Binary TreeGiven a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.递归和非递归，此提比较简单。广度优先遍历即可。关键之处就在于如何保持访问深度。下面是4种代码： 1import java.util.ArrayList;2import java.util.LinkedList;3import java.util.List;4import java.util.Queue;5 ...
题目描述：Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).For example:Given binary tree {3,9,20,#,#,15,7}, 3/ 9 20/ 15 7 return its level order traversal as:[[3],[9,20],[15,7] ]解题思路：运用广度优先搜索方法，同http://www.cnblogs.com/zihaowang/p/5149745.html类似。代码如下：/*** Definition for a binary tree node.* pu...
这是悦乐书的第285次更新，第302篇原创01 看题和准备今天介绍的是LeetCode算法题中Easy级别的第153题（顺位题号是671）。给定非空的特殊二叉树，其由具有非负值的节点组成，其中该树中的每个节点具有恰好两个或零个子节点。 如果节点具有两个子节点，则该节点的值是其两个子节点中的较小值。给定这样的二叉树，您需要输出由整个树中所有节点的值组成的集合中的第二个最小值。如果不存在这样的第二个最小值，则输出-1。例如： 2/...
A robot is located at the top-left corner of a m x n grid (marked ‘Start‘ in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish‘ in the diagram below). How many possible unique paths are there? 我的第一反应就是递归~~~每走一步下面的情况相当于变成m-1,n或者m，n-1的情形 也就是说：wa...
题目描述：Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree is symmetric: 1/ 2 2/ \ / 3 4 4 3 But the following is not: 1/ 2 2\ 3 3解题思路：递归法代码如下：/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode(int x) { v...
题目： Implement the following operations of a stack using queues.push(x) -- Push element x onto stack.pop() -- Removes the element on top of the stack.top() -- Get the top element.empty() -- Return whether the stack is empty.Notes:You must use only standard operations of a queue -- which means only pushto back, peek/pop from front, size,and is emptyoperations are valid.Depending on your language,...