双指针和滑动窗口怎么用？无重复字符的最长子串怎么做？

function maxArea(height) {
  let left = 0
  let right = height.length - 1
  let best = 0
  while (left < right) {
    const area = (right - left) * Math.min(height[left], height[right])
    best = Math.max(best, area)
    // 移动较短的一边：它和中间任何一条线组合都不会比现在大
    if (height[left] < height[right]) left++
    else right--
  }
  return best
}
console.log(maxArea([1, 8, 6, 2, 5, 4, 8, 3, 7])) // 49

left = 0
for right from 0 to n - 1:
    把 s[right] 加入窗口
    while 窗口需要收缩:
        把 s[left] 移出窗口，left++
    更新答案（求最长时在这里）

function lengthOfLongestSubstring(s) {
  const window = new Set() // 当前窗口里的字符
  let left = 0
  let best = 0
  for (let right = 0; right < s.length; right++) {
    const ch = s[right]
    // 加入 ch 会重复：收缩左边界，直到之前那个 ch 被移出窗口
    while (window.has(ch)) {
      window.delete(s[left])
      left++
    }
    window.add(ch)
    best = Math.max(best, right - left + 1) // 这时窗口一定合法
  }
  return best
}
console.log(lengthOfLongestSubstring('abcabcbb')) // 3
console.log(lengthOfLongestSubstring('')) // 0

function minWindow(s, t) {
  if (t.length === 0) return ''
  const need = new Map() // 字符 -> 还差几个，负数表示窗口里有富余
  for (const ch of t) need.set(ch, (need.get(ch) ?? 0) + 1)
  let missing = t.length
  let left = 0
  let bestStart = 0
  let bestLen = Infinity
  for (let right = 0; right < s.length; right++) {
    const ch = s[right]
    if (need.has(ch)) {
      if (need.get(ch) > 0) missing-- // 补上了一个缺的字符
      need.set(ch, need.get(ch) - 1)
    }
    while (missing === 0) {
      if (right - left + 1 < bestLen) [bestStart, bestLen] = [left, right - left + 1]
      const out = s[left++]
      if (need.has(out)) {
        need.set(out, need.get(out) + 1)
        if (need.get(out) > 0) missing++ // 移出的是必需字符，窗口不再合法
      }
    }
  }
  return bestLen === Infinity ? '' : s.slice(bestStart, bestStart + bestLen)
}
console.log(minWindow('ADOBECODEBANC', 'ABC')) // 'BANC'
console.log(minWindow('a', 'aa')) // ''