一.题目 Longest Substring Without Repeating Characters,具体请自行搜索. 这个题目,我看了一下,经过一番思考,我觉得实现起来不是很复杂. 但要做到bug free有点难度,主要是边界的问题. 二.这个题目,我自己实现,没有参考代码 提交了5次: 第1次: Wrong Answer,主要是" "这个空串不对 第2次.第3次:Runtime Error,"au" out of Range 第4次:Wrong Answer,…

题目: Given a string, find the length of the longest substring without repeating characters. Example 1: Input: "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3. Example 2: Input: "bbbbb" Output: 1 Exp…

Given a string, find the length of the longest substring without repeating characters. Example 1:           Input: "abcabcbb"                              Output: 3                           Explanation: The answer is "abc", with the l…

题目: 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 substring is “b”, with the…

题目等级:Medium 题目描述:   Given a string, find the length of the longest substring without repeating characters.   Example 1: Input: "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3.   Example 2: Input: "bbbbb…

题目如下: Given a string, find the length of the longest substring without repeating characters. Example 1: Input: "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3. Example 2: Input: "bbbbb" Output: 1 E…

题目: 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 s…

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters. For example, given:S: "b…

要求 在一个字符串中寻找没有重复字母的最长子串 举例 输入:abcabcbb 输出:abc 细节 字符集?字母?数字+字母?ASCII? 大小写是否敏感? 思路 滑动窗口 如果当前窗口没有重复字母,j右移,直到包含重复字母 i右移,直到不包含重复字母 用数组记录字母是否出现过,判断重复 实现 1 class Solution{ 2 public: 3 int lenthOfLongestSubstring(string s){ 4 int freq[256] = {0}; 5 int l = 0…

HashMap的应用可以提高查找的速度,键key,值value的使用拜托了传统数组的遍历查找方式,对于判断一个字符或者字符串是否已经存在的问题可以非常好的解决.而本题需要解决的问题就是判断新遍历到的字符是否已经存在于左left,右right,字符构成的子串之中. 解题思路:设置一个left作为子串的最左边的字符位置坐标,right不断向右遍历,并对每次遍历得到的字符进行判断,max作为最大的没有重复字符的子串长度.对于right遍历得到的新的字符,有两种情况: 一:在Hash表中没有,此时该字符…