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.
思路：
采用兩個(gè)變量left和res記錄最長(zhǎng)子串的左邊界和最長(zhǎng)子串的長(zhǎng)度嫩实。遍歷字符串裆站，如果沒(méi)有在hash表中出現(xiàn)過(guò)流昏，就更新長(zhǎng)度res舆声，如果出現(xiàn)過(guò)就更新左邊界的值。最后給每一位字符串對(duì)應(yīng)的位置填入hash表中（從1開(kāi)始）渔嚷。注意更新res長(zhǎng)度的判斷條件還有一個(gè)就是obj[s[i]]<left的值的時(shí)候即出現(xiàn)的這個(gè)字母即使存在過(guò)但他小雨左邊界进鸠，也需要更新長(zhǎng)度。
var lengthOfLongestSubstring = function(s) {
var obj = {};
var left = 0;
var res = 0;
for (var i = 0; i < s.length; i++) {
if (!obj[s[i]] || obj[s[i]] < left) {
res = Math.max(res, i - left + 1);
} else {
left=obj[s[i]];
}
obj[s[i]] = i + 1;
}
return res
};
