算法题:最长子序列问题_动态规划思路golang源码

package practices

var NUMS = []int{
   1, 5, 2, 3, 6, 9, 10, 4, 5, 55, 13, 12, 16, 17, 18, 20,
   18, 59, 2, 3, 6, 9, 10, 4, 5, 855, 13, 12, 16, 417, 18, 20,
   19, 500, 2, 3, 6, 9, 10, 4, 5, 755, 13, 212, 516, 517, 118, 120,
   100, 501, 42, 3, 6, 9, 610, 4, 611, 655, 616, 312, 116, 617, 118,
   18, 59, 2, 53, 6, 9, 10, 74, 5, 55, 513, 12, 416, 117, 178, 120,
   19, 500, 2, 63, 618, 9, 619, 84, 5, 620, 413, 12, 516, 217, 718, 120,
   100, 501, 2, 73, 6, 89, 10, 94, 5, 55, 133, 12, 616, 170, 718, 800,
}

func FindLongestSequence(nums []int, index int) int {
   // 定义出口
   if nums == nil {
      return 0
   }
   if index <= 0 {
      return 1
   }
   length := len(nums)
   if length == 0 || length == 1 {
      return length
   }
   // 比较当前第index的值和index-1的值的大小
   numBefore := nums[index-1]
   numThis := nums[index]
   if numBefore < numThis {
      // 如果当前值大，则最长子序列长度就是 前index-1个长度的子序列计算出的最大长度+1
      return 1 + FindLongestSequence(nums, index-1)
   } else {
      // 否则最大长度和index-1序列的最大长度相同
      return FindLongestSequence(nums, index-1)
   }
}

func main() {
   t1 := time.Now().UnixNano() / 1e6
   sum := practices.FindLongestSequence(practices.NUMS, len(practices.NUMS)-1)
   t2 := time.Now().UnixNano() / 1e6
   fmt.Println("===", sum, "\nt1=", t1, "\nt2=", t2, "\ntimeCost:=", t2-t1)
}

执行结果如下：

这个问题很简单，源代码注释写的有每一步骤的思路。每一个子序列的最大长度只用计算一次。