Just DO IT by luckygxf

摘要： 1 package com.gxf.search; 2 3 /** 4 * 测试折半查找or二分查找 5 * @author xiangfei 6 * 7 */ 8 public class BiSearch { 9 10 /**11 * 非递归实现,从第1个元... 阅读全文

摘要： Climbing StairsYou are climbing a stair case. It takesnsteps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct way... 阅读全文

摘要： 静态查找设置哨兵，可以减少比较次数，效率会有所提高。静态查找算法简单，效率较低 1 package com.gxf.search; 2 3 /** 4 * 静态查找实现 5 * @author xiangfei 6 * 7 */ 8 public class StaticSearch { ... 阅读全文

摘要： Length of Last WordGiven a stringsconsists of upper/lower-case alphabets and empty space characters' ', return the length of last word in the string.I... 阅读全文

摘要： Dijkstra算法又称单源点距离最短问题设一个图中有V0,V1,V2,V3...等顶点，这里设求的是V0到V1,V2,...的最短距离基本思想V0到剩余顶点的直接距离dist[]（不经过其他任何节点，没有联通的设为无穷大）中,找出一个最小的，设其顶点为V1，这里我们就求出了V0到V1的最短距离。将... 阅读全文

摘要： Valid ParenthesesGiven a string containing just the characters'(',')','{','}','['and']', determine if the input string is valid.The brackets must clos... 阅读全文

摘要： 感觉和c++的模板很类似，这里记录一下基本的认识 1 public class Test { 2 private T obj; 3 public Test(T obj){ 4 this.obj = obj; 5 } 6 public T getObj(... 阅读全文

摘要： Remove Duplicates from Sorted ArrayGiven a sorted array, remove the duplicates in place such that each element appear onlyonceand return the new lengt... 阅读全文

摘要： Add BinaryGiven two binary strings, return their sum (also a binary string).For example,a ="11"b ="1"Return"100". 1 public class Solution { 2 publ... 阅读全文

摘要： Plus OneGiven a non-negative number represented as an array of digits, plus one to the number.The digits are stored such that the most significant dig... 阅读全文