3. Longest Substring Without Repeating Characters
Tags
- String
- Hash Table
- Two Pointers
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.
题意：
给定一个字符串，寻找字符串中不重复最长子串的长度
分析：
分析题目，抓住两个关键词：
- 最长子串 =>寻找子串起点和终点位置
- 不重复 =>循环比较字符是否重复
思路：
由此，第一个想法就是逐个字符比对，暴力搜索。但是如果完全循环的话，时间复杂度太高，将达到O(n3);肯定会出现Time Limit Exceeded。因此想到了做改良。思路如下：
- 将字符串第i位与其之前的每位字符一一比较
- 如果不相等，记录两者位置距离长度
- 如果相等，退出，返回之前记录的距离长度
- 注意一开始就重复的特殊情况‘aa’
Js实现
方法1是自己的实现，后面几个方法都是谷歌后使用JavaScript实现的
方法 #1 暴力搜索
复杂度
时间 <= O(n2) && >= O(n)
let lengthOfLongestSubstring = function(s) {
let index = 0;
let maxstr = 0;
let len = s.length;
if (len === 0 || s === null) return 0;
if (len === 1) return 1;
for (let i = 1; i < len; i++) {
for (let j = i - 1; j >= index; j--) {
if (s[i] === s[j]) {
index = j + 1;
if (i == 1) {
maxstr = 1;
}
break;
} else {
if (maxstr < i - j + 1)
maxstr = i - j + 1;
}
}
}
return maxstr;
};
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方法 #2 滑动窗口
复杂度
时间O(n)
let lengthOfLongestSubstring = function(s) {
let n = s.length;
let set = new Set();
let ans = 0, i = 0, j = 0;
while (i < n && j < n) {
// try to extend the range [i, j]
if (!set.has(s.charAt(j))){
set.add(s.charAt(j++));
ans = Math.max(ans, j - i);
}
else {
set.delete(s.charAt(i++));
}
}
return ans;
}
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方法 #3 扩展的ASCII码 Map存储
复杂度
时间O(n)
let lengthOfLongestSubstring = function(s) {
let map = new Map(); // map from character's ASCII to its last occured index
let ans = 0;
let slow = 0;
for(let i =0;i<256;i++) map.set(i,-1);
for (let fast = 0; fast < s.length; fast++) {
let ch = s.charAt(fast);
if (map.get(ch) >= slow) {
ans = Math.max(ans, fast - slow);
slow = map.get(ch) + 1;
}
map.set(ch,fast);
}
return Math.max(ans, s.length - slow);
}
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方法 #4 扩展的ASCII码 Table数组存储
复杂度
时间O(n)
let lengthOfLongestSubstring = function(s) {
let start = 0; // current start point of substring without dup
let maxlen = 0; // max length of substring found
let table= new Array(256); // hash table for index of each char appeared
for (let i = 0;i < 256;i++) table[i] = -1; // if char not present, index is -1
let len = s.length;
for (let i = 0;i < len;i++) {
//console.log(table[s[i]]);
if (table[s[i].codePointAt(0)] != -1) {
while (start <= table[s[i].codePointAt(0)]) table[s[start++].codePointAt(0)] = -1;
}
if (i - start + 1 > maxlen) maxlen = i - start + 1;
table[s[i].codePointAt(0)] = i;
}
return maxlen;
}
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方法 #5 滑动窗口 使用Map
复杂度
时间O(n)
let lengthOfLongestSubstring = function(s) {
let n = s.length, ans = 0;
let map = new Map(); // current index of character
// try to extend the range [i, j]
for (let j = 0, i = 0; j < n; j++) {
if (map.has(s.charAt(j))) {
i = Math.max(map.get(s.charAt(j)), i);
}
ans = Math.max(ans, j - i + 1);
map.set(s.charAt(j), j + 1);
}
return ans;
}
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方法 #6 ASCII 128
复杂度
时间O(n)
let lengthOfLongestSubstring = function( s) {
let n = s.length, ans = 0;
let index = new Array(128); // current index of character
for(let i = 0; i < 128;i++) index[i] = -1;
// try to extend the range [i, j]
for (let j = 0, i = 0; j < n; j++) {
i = Math.max(index[s.charAt(j).codePointAt(0)], i);
ans = Math.max(ans, j - i + 1);
index[s.charAt(j).codePointAt(0)] = j + 1;
}
return ans;
}
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