/**
 * @param {number[]} nums
 * @return {number}
 */
let lengthOfLIS = function (nums) {
  let dp = []
  let n = nums.length
  if (!n) {
    return 0
  }

  dp[0] = 1
  for (let i = 1; i < n; i++) {
    let num = nums[i]
    let max = 1
    // j 从 [0, i) 依次求出可以和 i 组成的最长上升子序列
    for (let j = 0; j < i; j++) {
      let prevNum = nums[j]
      if (num > prevNum) {
        // 循环中不断更新 max 值
        max = Math.max(max, dp[j] + 1)
      }
    }
    dp[i] = max
  }

  return Math.max(...dp)
}