Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest subst
# encoding:utf-8 # p001_1234threeNums.py def threeNums(): '''题目:有1.2.3.4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少?''' print None count = 0 nums = [] for index1 in xrange(1,5): for index2 in xrange(1,5): for index3 in xrange(1,5): if index1 != index2 and index1 !
有一类面试题,既可以考察工程师算法.也可以兼顾实践应用.甚至创新思维,这些题目便是好的题目,有区分度表现为可以有一般解,也可以有最优解.最近就发现了一个这样的好题目,拿出来晒一晒. 1 题目 原文: There is an array of 10000000 different int numbers. Find out its largest 100 elements. The implementation should be optimized for executing speed. 翻译
题目 最长无重复字符的子串给定一个字符串,请找出其中无重复字符的最长子字符串. 例如,在"abcabcbb"中,其无重复字符的最长子字符串是"abc",其长度为 3. 对于,"bbbbb",其无重复字符的最长子字符串为"b",长度为1. 解题 利用HashMap,map中不存在就一直加入,存在的时候,找到相同字符的位置,情况map,更改下标 public class Solution { /** * @param s: a s
大一学期末的时候做课程设计时遇到过生成无重复随机数的问题,今天自己也写出来了: static int[] Create_Value() { Random ran = new Random(); //生成0-51之间的无重复随机数,作为纸牌数组的索引 int[] a = new int[52]; for (int j = 0; j < 52; j++) { again: int x = ran.Next(52); a[j] = x; for (int m = 0; m < j; m++) { i
Given a string, find the length of the longest substring without repeating characters. Examples: Given "abcabcbb", the answer is "abc", which the length is 3. Given "bbbbb", the answer is "b", with the length of 1.
