Longest Common Substring 最长公共子字符串 动态规划问题 动态规划问题的两个特点: 1.最优子结构 2.重叠子问题 因为有重叠子问题,当前计算的过程中可能有的问题在之前的计算已经计算过了,现在又要计算一遍,导致大量重复的计算。 动态规划通过找到解决问题的递推关系,将已经完成计算的存储起来, 当开始新的计算时如果...
Given two stringstext1andtext2, return the length of their longest common subsequence. Asubsequenceof a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, "ace" is a subsequen...
5. Longest Common Substring,最长子字符串系列,13个题 Longest Common Substring,最长相同子串 Longest Common Subsequence,最长相同子序列 Minimum Deletions & Insertions to Transform a String into another,字符串变换 Longest Increasing Subsequence,最长上升子序列 Maximum Sum Increasing Subsequence,最长上升子序列和 ...
最长公共子串LCS(Longest Common Substring) 一、问题描述 寻求两个字符串中的最大公共字串,其中子串是指字符串中连续的字符组成的,而不是像子序列,按照字符的前后顺序组成。 如str1="sgabacbadfgbacst",str2="gabadfgab",则最长公共字串为"badfg",这些字符是连续的. 二、问题分析 对字符串a中的每个字符...
可以在地址 https://en.wikipedia.org/wiki/Longest_common_substring_problem 阅读更多关于最长公共子串的内容。 此处便不再赘述了! 4解法二 暴力法 选出所有子字符串可能的开始和结束位置,并检验它是不是回文。 当字符串中字符出现次数为偶数时,必然可以加入最长回文子串 ...
Given a stringS, find the longest palindromic substring inS. You may assume that the maximum length ofSis 1000, and there exists one unique longest palindromic substring. http://www.acmerblog.com/longest-palindromic-subsequence-5721.html
Leetcode 1143 Longest Common Subsequence (前缀型dp) Leetcode 1062 Longest Repeating Substring (dp方法与longest common substring一致) Leetcode 718 Maximum Length of Repeated Subarray (和1062本质上一样) Leetcode 174 Dungeon Game Leetcode 115 Distinct Subsequences Leetcode 72 Edit Distance Leetcode 91...
public String longestCommonPrefix(String[] strs) { if (strs == null || strs.length == 0) return ""; String prefix = strs[0]; for (int i = 1; i < strs.length; i++) { while (strs[i].indexOf(prefix) != 0) {
1. public String longestCommonPrefix(String[] strs) { 2. if (strs.length == 0) { 3. return ""; 4. } 5. int miniLength = strs[0].length(); 6. for (String s : strs) { 7. if (s.length() < miniLength) { 8. miniLength = s.length(); ...
{ int mid = (end - start) / 2 + start; String lcpLeft = longestCommonPrefix(strs, start, mid); String lcpRight = longestCommonPrefix(strs, mid + 1, end); return commonPrefix(lcpLeft, lcpRight); } } public String commonPrefix(String lcpLeft, String lcpRight) { int minLength = ...