题目: Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a…

72. Edit Distance Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) You have the following 3 operations permitted on a word: a) Insert a character b) Delete a…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 记忆化搜索 动态规划 日期 题目地址:https://leetcode.com/problems/edit-distance/description/ 题目描述 Given two words word1 and word2, find the minimum number of operations required to convert w…

[题解]UVA10140 Prime Distance 哈哈哈哈\(miller-rabbin\)水过去了哈哈哈 还能怎么办呢?\(miller-rabbin\)直接搞.枚举即可,还跑得飞快. 当然此题由于\(20000^2 >2^{31}\),直接预处理\(20000\)内的质数就好了 放mr的代码 #include<bits/stdc++.h> using namespace std;typedef long long ll; #define DRP(t,a,b) for(regis…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/maximize-distance-to-closest-person/description/ 题目描述 In a row of seats, 1 represents a person sitting in that seat, and 0 represents that…

 二分法:通过O(1)的时间,把规模为n的问题变为n/2.T(n) = T(n/2) + O(1) = O(logn). 基本操作:把长度为n的数组,分成前区间和后区间.设置start和end下标.int start = 0, end = nums.length - 1.循环结束条件为start + 1 < end ,即相邻时结束循环,所以最后需判断start和end中哪个是目标值.指针变化为start = mid,取后半区间:end = mid,取前半区间. 经典二分搜索:1. First p…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given two words word1 and word2, find the minimum number of steps ?>required to convert word1 to word2. (each operation is counted as 1 step.) You have the f…

题意:给定一个长为n的序列,有m次强制在线的询问,每次询问位置[L,R]中abs(a[i]-p)第k小的值 n,m<=1e5,a[i]<=1e6,p<=1e6,k<=169 思路:主席树外面套个二分 #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned int uint; typedef unsigned long long ull; typedef pai…

problem 461. Hamming Distance solution1: 根据题意,所求汉明距离指的是两个数字的二进制对应位不同的个数.对应位异或操作为1的累积和. class Solution { public: int hammingDistance(int x, int y) { ; ; i<; i++) { <<i))^(y&(<<i))) ans++; } return ans; } }; solution2: 两个数字异或之后,统计结果二进制中1的…

题目: The code base version is an integer start from 1 to n. One day, someone committed a bad version in the code case, so it caused this version and the following versions are all failed in the unit tests. Find the first bad version. You can call isBa…