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.

Given “pwwkew”, the answer is “wke”, with the length of 3. Note that the answer must be a substring, “pwke” is a subsequence and not a substring.

0x01 分析

目的是想要截取到给定字符串中最长的不重复的字符串。

eg:
abcabcabc => abc

aaaaaaaababcd => abcd

aaaaaa => a

0x02 编码

• golang 版本

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  func lengthOfLongestSubstring(s string) int { location := make([]int, 128) for i := 0; i < 128; i++ { location[i] = -1 } begin, length := -1, 0 for i, v := range s { if location[v] > begin{ begin = location[s[i]] } if (i - begin > length) { length = i - begin; } location[s[i]] = i; } return length }

0x03 运行结果

传送门：Longest Substring Without Repeating Characters

时隔一年，重新写起博客，以前的东西没有存，随它去吧，重新开始。
去年年初接触到 LeetCode，这是一个Online Judge网站，很有趣，从这篇博客开始，争取把所有题目都做完。

0x00 题目

Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.

You may assume the integer do not contain any leading zero, except the number 0 itself.

The digits are stored such that the most significant digit is at the head of the list.

0x01 分析

根据字面意思简单的理解就是给一个正整数组成的数组，期望将数组中的数字看成一个整型数，对其进行加一操作。
eg:
[1] => [2]
[9] => [1, 0]
[1, 9, 9] => [2, 0, 0]

0x02 编码

• python 版本

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  class Solution(object): def plusOne(self, digits): if len(digits) == 0: return digits if digits[0] == 9 and len(digits) == 1: return [1,0] if digits[-1] != 9: digits[-1] += 1 return digits tmp = self.plusOne(digits[0:-1]) tmp.append(0) return tmp

0x03 运行结果

传送门：Plus One