【Leet Code】Longest Substring Without Repeating Characters

標籤：遍曆   substring   

Longest Substring Without Repeating Characters Total Accepted: 20506 Total Submissions: 92223My Submissions

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.

這題目有點兒坑啊，一開始沒想明白，過一會兒才看懂題目要求，題目是找沒有重複字元的最長子字串，呵呵。也沒想到什麼好辦法，只能while了：

class Solution {public:int lengthOfLongestSubstring(string s) {bool state[256];memset(state, false, sizeof(state));int ans = 0, l = 0, r = 0;while (r < s.length()) {while (r < s.length() && state[ s[r] ] == false) {state[ s[r++] ] = true;}ans = max(ans, r - l);while (l < r && s[l] != s[r]){state[s[l++]] = false;}l++, r++;}return ans;}};

這個方法很明顯了，用while裡面嵌套while，可以說遍曆的大部分字串了。不過，有大神的代碼如下：

class Solution {public:int pos[256];int lengthOfLongestSubstring(string s){// IMPORTANT: Please reset any member data you declared, as// the same Solution instance will be reused for each test case.for (int i = 0; i < 256; ++i)pos[i] = -1;int stp = -1, sz = s.size(), res = 0;for (int i = 0; i < sz; ++i){if (pos[s[i]] >= stp){  //update posistionstp = pos[s[i]] + 1;}pos[s[i]] = i;res = max(res, i - stp + 1);}return res;}};//whose code is shorter than mine? Please notify me! I want to meet shorter codes!

大神就是大神，居然把找到的最近重複的字元的位置給記錄下來了，佩服佩服，更讓人為之瘋狂的是，居然還在最後加上了：

<span style="font-size:18px;color:#ff0000;"><strong>whose code is shorter than mine? Please notify me! I want to meet shorter codes!</strong></span>

哈哈，厲害厲害！

【Leet Code】Longest Substring Without Repeating Characters