015-无重复字符的最长子串 (Longest Substring Without Repeating Characters)

输入: s = "abcabcbb"
输出: 3
解释: 因为无重复字符的最长子串是 "abc"，所以其长度为 3。

输入: s = "bbbbb"
输出: 1
解释: 因为无重复字符的最长子串是 "b"，所以其长度为 1。

输入: s = "pwwkew"
输出: 3
解释: 因为无重复字符的最长子串是 "wke"，所以其长度为 3。
请注意，你的答案必须是 子串 的长度，"pwke" 是一个子序列，不是子串。

/**
 * @param {string} s
 * @return {number}
 */
const lengthOfLongestSubstring = (s) => {
    const charMap = new Map();
    let maxLength = 0;
    let left = 0;
    
    for (let right = 0; right < s.length; right++) {
        const currentChar = s[right];
        
        // 如果字符已经在窗口中，更新左边界
        if (charMap.has(currentChar) && charMap.get(currentChar) >= left) {
            left = charMap.get(currentChar) + 1;
        }
        
        // 更新字符的最后出现位置
        charMap.set(currentChar, right);
        
        // 更新最大长度
        maxLength = Math.max(maxLength, right - left + 1);
    }
    
    return maxLength;
};

// 测试函数
const testLengthOfLongestSubstring = () => {
    const testCases = [
        {
            s: "abcabcbb",
            expected: 3
        },
        {
            s: "bbbbb",
            expected: 1
        },
        {
            s: "pwwkew",
            expected: 3
        },
        {
            s: "",
            expected: 0
        },
        {
            s: "a",
            expected: 1
        },
        {
            s: "ab",
            expected: 2
        },
        {
            s: "dvdf",
            expected: 3
        },
        {
            s: "anviaj",
            expected: 5
        },
        {
            s: "abcabcbb",
            expected: 3
        },
        {
            s: "tmmzuxt",
            expected: 5
        }
    ];

    testCases.forEach((testCase, index) => {
        const result = lengthOfLongestSubstring(testCase.s);
        const passed = result === testCase.expected;
        console.log(`测试用例 ${index + 1}: ${passed ? '通过' : '失败'}`);
        console.log(`输入: s = "${testCase.s}"`);
        console.log(`期望: ${testCase.expected}, 实际: ${result}\n`);
    });
};

// 运行测试
testLengthOfLongestSubstring();