js 写一个DC3算法创建后缀数组

好的，我可以辅助您回答问题。以下是利用 Javascript 语言编写 DC3 算法创建后缀数组的代码：

// 定义用于 DC3 算法排序的变量
let t1, t2, sa12, sa0

function sortRank(t, sa, n, m) {
  // 记录排序结果到 t 数组

  for (let i = 0; i < m; i++) counter[i] = 0
  for (let i = 0; i < n; i++) counter[t[sa[i]]]++
  let sum = 0
  for (let i = 0; i < m; i++) {
    let t = counter[i]
    counter[i] = sum
    sum += t
  }
  for (let i = 0; i < n; i++) {
    let k = sa[i]
    t2[counter[t[k]]++] = k
  }

  for (let i = 0; i < m; i++) counter[i] = 0
  for (let i = 0; i < n; i++) counter[t[t2[i] + k < n ? sa[t2[i] + k] : 0]]++
  sum = 0
  for (let i = 0; i < m; i++) {
    let t = counter[i]
    counter[i] = sum
    sum += t
  }
  for (let i = n - 1; i >= 0; i--) {
    let k = t2[i]
    sa[--counter[t[k + k < n ? sa[k + k] : 0]]] = k
  }

  // 合并排序结果，保存到 sa 数组
  let j = 0
  for (let i = 0; i < n; i++) {
    let p = sa[i], q = sa[j]
    let pp = p + k < n ? t[p + k] : -1, qq = q + k < n ? t[q + k] : -1
    if (pp != qq) {
      pp = -1
    }
    if (p + k < n && pp == qq) {
      if (q < sa12n) {
        pp = t[p + k + sa12n] == t[q + k + sa12n] ? 1 : 2
      } else {
        pp = t[p + k] < t[q + k] ? 1 : 2
      }
    }
    sa[j++] = pp == 1 ? p : q
  }
}

function createSa(t, sa, n, m) {
  sa12 = new Array(n / 3 + 1)
  sa0 = new Array(n / 3)
  let n0 = (n + 2) / 3
  let n12 = n / 3 * 2 - (n % 3 == 1 ? 1 : 0)

  // 第一次基数排序
  let j = 0
  for (let i = 0; i < n + n0 - n12; i++) {
    if (i % 3 == 1) {
      t1[j++] = i
    }
  }
  sortRank(t, t1, j, m)
  for (let i = 0; i < j; i++) {
    sa12[t1[i] / 3] = i + 1
  }
  if (j + 1 == n12) {
    sa12[n12 / 3] = 0
  }

  // 对 12 数组排序
  sortRank(t, sa12, n12, m)

  // 记录 12 数组对应的下标，便于第二次比较
  for (let i = 0, k = 0; i < n12; i++) {
    let x = sa12[i] * 3 - 2
    if (x < n) {
      t2[k++] = x
    }
  }
  for (let i = 0, k = 0; i < n; i++) {
    if (i % 3 != 1) {
      t2[k++] = i
    }
  }

  // 第二次基数排序
  sortRank(t, t2, n12, m)

  // 合并各个字符对应的后缀
  let p = 0
  for (let i = 0, j = n12 - n0, k = 0; k < n; k++) {
    let pp = sa12[j] * 3 + (i % 3 == 1 ? 0 : 1)
    let qq = t2[i]
    if (j == n12 || (i < n0 && t[pp] < t[qq + k < n ? sa12[j + n0] * 3 + (qq + k < n ? 0 : 2) : 0])) {
      sa[k] = pp
      i++
    } else {
      sa[k] = qq
      j++
    }
  }
}

function dc3(t, n) {
  n++
  t1 = new Array(n), t2 = new Array(n)
  let n0 = (n + 2) / 3
  let ns = (n + 1) / 3
  let n2 = n / 3
  let n02 = n0 + n2
  let sa = new Array(n02 + 3)

  // 复制 t，将其长度扩充到 3 的倍数
  let tCopy = new Array(n02 + 3)
  for (let i = 0, j = 0; i < n02 - 1; i++) {
    tCopy[i] = i % 3 == 0 ? 0 : t[j++]
  }
  tCopy[n02 - 1] = 0

  // 创建后缀数组
  createSa(tCopy, sa, n02 - 1, 128)
  sa[n02] = sa[n02 + 1] = sa[n02 + 2] = 0

  // 接下来要用的排序变量
  let k = 0
  let s = 0
  let p = n0 - ns
  let sa12n = n02 - n0

  // 将 12 数组去重，增加索引对应下标
  for (let i = 0; i < n02; i++) {
    let v = sa[i]
    if (v < n0 || (v >= n02)) {
      continue
    }
    v -= n0
    t2[k++] = v * 3 + 1
    t2[k++] = v * 3 + 2
  }
  sortRank(t, t2, k, n02)

  // 将去重后的 12 数组重新映射到 sa12
  for (let i = 0; i < k; i++) {
    let v = t2[i]
    sa12[s = v / 3 + (v % 3 == 1 ? p : 0)] = i + 1
  }

  // 如果 12 数组去重后还有重复值，递归进行 DC3 算法
  if (s + 1 < sa12n) {
    dc3(t2, sa12n)
    for (let i = 0; i < n02; i++) {
      sa12[t2[i] < n0 ? t2[i] * 3 + 1 : (t2[i] - n0) * 3 + 2] = i + 1
    }
  } else {
    for (let i = 0; i < sa12n; i++) {
      sa[t2[i] - n0] = i + 1
    }
  }

  // 将各个字符对应的后缀排序
  k = 0
  for (let i = 0; i < n02; i++) {
    let v = sa[i]
    if (v < n0) {
      t2[k++] = v * 3
    } else if (v < n02) {
      t2[k++] = (v - n0) * 3 + 1
    } else {
      t2[k++] = (v - n02) * 3 + 2
    }
  }
  sortRank(t, t2, k, n02)

  // 将 t2 中排序结果赋值到 sa 数组
  for (let i = 0; i < n02; i++) {
    sa[i] = 0
  }
  let j = 0
  for (let i = 0; i < k; i++) {
    let v = t2[i]
    if (i == 0 || t[v] != t[t2[i - 1]]) {
      sa[j++] = v / 3
    }
  }
  for (let i = 0; i < n0; i++) {
    sa[j++] = i
  }

  // 进行最后一次排序
  for (let i = 0; i < n02; i++) {
    t1[sa[i]] = i + 1
  }
  for (let i = 0; i < n02; i++) {
    sa[i] = sa[i] < n0 ? sa[i] * 3 + 1 : (sa[i] - n0) * 3 + 2
  }
  sortRank(t, sa, n02, n)

  // 返回后缀数组
  return sa
}

// 调用示例
let str = "banana"
let t = str.split("").map((c) => c.charCodeAt(0))
let sa = dc3(t, t.length)
console.log(sa)

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