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输出S中的最长回文串eg.Input: s = "babad"Output: "bab"Approach 1纯暴力匹配，提交超时Approcah 2 Manacher算法将暴力匹配的算法复杂度提升到O(n)，先对S进行处理，每个字符前后插入"#"(或其他S内不可能出现的字符)，经过处理的S'长度一定为奇数，分别以每一个字符为中心计算他的回文半径len，原字符串中它的回文长度为len-1，所以问题变为求每一个字符的len[i]。p:最右回文边界center：当前p的中心.

1、介绍 动态规划是通过组合子问题的解而解决整个问题，动态规划适用于子问题不是独立的情况，也就是各子问题包含公共的子子问题。动态规划算法对每个子子问题只求解一次，将其结果保存在一张表中，从而避免每次遇到各个子问题时重新计算。 动态规划通常应用于最优化问题，此类问题可能有很多种可行解。每个解有一个值，而我们希望找出一个具有最优（最大或最小）值的解。称这样的解为该问题的“一个”最优解（而不是“确定”的“最优解”），因为可能存在多个取最优值的解。 动态规划算法...

题目：Given anon-emptyarray of decimal digitsrepresenting a non-negative integer, incrementone to the integer.The digits are stored such that the most significant digit is at the head of the list, and each element in the array contains a single digit....

题目：Given an integer arraynums, find the contiguous subarray (containing at least one number) which has the largest sum and returnits sum.思路：返回由相邻数组相加得到的最大的和class Solution { public int maxSubArray(int[] nums) { int sum = nums[0]; in..

题目：Merge two sorted linked lists and return it as asortedlist. The list should be made by splicing together the nodes of the first two lists.将两个ListNode按照顺序拼接成一个，一开始想通过l1.val和l2.val比较大小，再借助ListNode temp将其插入至其中一个ListNode中，但在一开始调试过程中发现，若temp=q；（q=l2...

题目：Write a function to find the longest common prefix string amongst an array of strings.If there is no common prefix, return an empty string"".思路：要求输出strs中最长重复的字符串，使用String类中的compareTo()方法比较字符串，使用substring（int beginIndex,int endIndex）截取字符串。该方法判断较.

题目：You are given twonon-emptylinked lists representing two non-negative integers. The digits are stored inreverse order, and each of their nodes contains a single digit. Add the two numbers and return the sumas a linked list.You may assume the two ...

题目：Given an array of integers, return indices of the two numbers such that they add up to a specific target.You may assume that each input would have exactly one solution, and you may not use the ...