Leetcode 3.无重复字符的最长子串 JavaScript (Day 4)

js一刷法一(布尔集合)

var lengthOfLongestSubstring = function (s) {
    let idx = new Set();
    let i = 0;
    let len = 0;

    for (let j = 0; j < s.length; j++) {


        while (idx.has(s[j])) {
            idx.delete(s[i]);
            i++;
        }

        idx.add(s[j]);
        len = Math.max(len, j - i + 1);
    }

    return len;
};

js一刷法二

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function (s) {
    let idx = new Map();
    let i = 0;
    let len = 0;

    for (let j = 0; j < s.length; j++) {


        while (idx.get(s[j])) {
            idx.set(s[i], idx.get(s[i]) - 1);
            i++;
        }

        idx.set(s[j], (idx.get(s[j]) ?? 0) + 1);
        len = Math.max(len, j - i + 1);
    }

    return len;
};

核心思想:滑动窗口
语法细节: (idx.get(s[j]) ?? 0),如果左边是false的话返回0,否则返回idx.get(s[j])