[LeetCode-interview algorithm classic-Java implementation] [003-Longest Substring Without Repeating Characters (Longest non-duplicate Substring)], longestsubstring

[LeetCode-interview algorithm classic-Java implementation] [003-Longest Substring Without Repeating Characters (Longest non-duplicate Substring)], longestsubstring
[003-Longest Substring Without Repeating Characters (maximum non-duplicate Substring )]Original question

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 length of 1.

Theme

Returns the maximum number of non-duplicate substrings in a string.

Solutions

Use start to record the start position of the current processing string. When the current character has appeared since start, the start position of the substring is + 1, otherwise, the hash value in the map is updated to the current position.

Code Implementation

Import java. util. arrays; import java. util. hashMap; import java. util. map;/*** Author: Wang junchao * Date: 2015-06-17 * Time: 20:46 * Declaration: All Rights Reserved !!! */Public class Solution {/*** 003-Longest Substring Without Repeating Characters (maximum non-duplicate Substring) ** @ param s input String * @ return maximum length of a non-duplicate substring * // can process all UTF-8 characters public int lengthOfLongestSubstring (String s) {// The string input is invalid if (s = null) {return 0;} // the start position of the current processing int start = 0; // The maximum length of the non-duplicate substring recorded int result = 0; // access tag, record the last accessed Character and position Map <Character, integer> map = new HashMap <> (s. length (); for (int I = 0; I <s. length (); I ++) {char ch = s. charAt (I); // if the character already exists (starting from the starting position), remark start if (map. containsKey (ch) & map. get (ch)> = start) {start = map. get (ch) + 1 ;}// if no such occurrence occurs, calculate the maximum length of the non-duplicate substring else {result = Math. max (result, I-start + 1);} // update the access record map. put (ch, I);} return result;} // only ASCII characters are considered. [solution 2] public int lengthOfLongestSubstring2 (String s) {// The string input is invalid if (s = null) {return 0;} // indicates whether the mark character exists, and the location of the last accessed element is recorded as int [] appear = new int [256]; // It is initialized to-1 Arrays. fill (appear,-1); // start position of the current processing int start = 0; // Save the result int result = 0; for (int I = 0; I <s. length (); I ++) {// if the character already exists (in the marked open position), re-mark start if (appear [s. charAt (I)]> = start) {start = I + 1;} // if there is no such occurrence, calculate the maximum length of the non-duplicate substring else {result = Math. max (result, I-start + 1);} // mark that the I character has been accessed (the latest is the I position) appear [s. charAt (I)] = I;} return result ;}}

Evaluation Result

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Note Please refer to the following link for more information: http://blog.csdn.net/derrantcm/article/details/46921991]

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