标签：turn out string 分享 ram property 遍历 col new
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.
给定一个字符串，找到最长的子串的长度没有重复字符。
这题其实思路就是把字符串中的字符遍历，记录每个字符的位置，遇到相同的字符更新位置，然后比较长度。可是自己做的乱七八糟，多出很多没用的东西。
/** * @param {string} s * @return {number} */ var lengthOfLongestSubstring = function(s) { var count = 0,//本来想用来记录当前连续无重复字符数，但其实不需要，i-start+1就等于它 obj = {}, arr = s.split(‘‘), start = 0, length = 0, del = "",//本来想用来记录，然后删除重复字符之前的，没用了的字符，但其实完全没必要，只需要记录历史以前最长长度length和当前无重复字符开始位置start就可以 Substring = "";//审题不仔细，本来以为要返回子字符串的 for(var i=0;i<arr.length;i++){ //debugger; if(obj[arr[i]] === undefined){ }else{ start = obj[arr[i]] + 1; } if(i-start+1 > length){ Substring = s.slice(start,i+1); } obj[arr[i]] = i; } return length; };
然后对比热门答案，真是无地自容，解题方法没想到都还算了，本质思路其实不差多少，却多出来那么多垃圾代码。以后要注意审视自己写的东西了
public int lengthOfLongestSubstring(String s) { if (s.length()==0) return 0; HashMap<Character, Integer> map = new HashMap<Character, Integer>(); int max=0; for (int i=0, j=0; i<s.length(); ++i){ if (map.containsKey(s.charAt(i))){ j = Math.max(j,map.get(s.charAt(i))+1); } map.put(s.charAt(i),i); max = Math.max(max,i-j+1); } return max; }
var lengthOfLongestSubstring = function(s) { var obj = {}, arr = s.split(‘‘), start = 0, length = 0; for (var i = 0; i < arr.length; i++) { if (obj.hasOwnProperty(arr[i])) { start = obj[arr[i]] + 1； } obj[arr[i]] = i; length = Math.max(length, i - start + 1) } return length; };
LeetCode解题笔记 - 3. Longest Substring Without Repeating Characters
标签：turn out string 分享 ram property 遍历 col new
原文地址：http://www.cnblogs.com/liyan-web/p/7857665.html
