題目
Given a string, find the length of the longest substring without repeating characters.
example
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.
分析
這是leetcode上的第3題，難度為medium，就是求一字符串的連續(xù)子串長度，該子串中不存在重復(fù)字符。我的思路是先把字符串轉(zhuǎn)成數(shù)組，同時用一個臨時數(shù)組存放子串。在遍歷數(shù)組時，如果當前元素在臨時數(shù)組中已經(jīng)存在，記錄此時子串的最大長度，再把已經(jīng)存在的元素及之前的元素移出數(shù)組，最后把當前元素push到臨時數(shù)組中，直至遍歷完成。
js實現(xiàn)
/**
* @param {string} s
* @return {number}
*/
var lengthOfLongestSubstring = function(s) {
var arr = s.split('');
var longest = 0;
var temp = [];
arr.forEach(function(value){
let index_of = temp.indexOf(value);
if(index_of!=-1){//當前元素已存在
longest = temp.length > longest ? temp.length : longest;
for(let i = 0;i <= index_of;i++){
temp.shift();
}
}
temp.push(value);
});
return temp.length > longest ? temp.length : longest;
};
總結(jié)
討論區(qū)有個大神用c寫的代碼執(zhí)行時間僅為4ms，而上面的js代碼執(zhí)行時間為185ms，他是用指針記錄子串的開始和結(jié)束，然后對上面代碼進行改進，如下：
/**
* @param {string} s
* @return {number}
*/
var lengthOfLongestSubstring = function(s) {
var arr = s.split('');
var longest = 0;
var start = 0,end = -1;
arr.forEach(function(value){
let index_of = arr.indexOf(value,start);//不能從頭開始遍歷
if(index_of >= start && index_of <= end){//當前元素已存在
longest = end -start +1 > longest ? end -start +1 : longest;
start = index_of + 1;
}
end++;
});
return end -start +1 > longest ? end -start +1 : longest;
};
運行時間182ms，好像優(yōu)化不明顯，有興趣的可以試試再優(yōu)化一下~
