力扣题目汇总（最长连续递增序列，旋转图像（中等），宝石与石头）

如果后者大于前者长度加1，后者小于前者长度变回1，然后重新开始循环如果后者大于前者加1

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        new_nums = nums[1:]
        count = 1
        len_list = []
        if len(nums)==0:
            return 0
        else:
            for a in range(len(new_nums)):
                if new_nums[a] > nums[a]:
                    count += 1
                else:
                    len_list.append(count)
                    count = 1
            len_list.append(count)                
            return max(len_list)       

给定 matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

原地旋转输入矩阵，使其变为:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

给定 matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

原地旋转输入矩阵，使其变为:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

难点：原表格修改，用太多for循环会超时
先水平翻转，再按照子列表的长度，把他变成有拥有子列表长度一样的空的列表，再里面填写值进去

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        import copy
        new_list = []
        time = 0
        conter = 0
        matrix.reverse()
        matrix_1 = copy.copy(matrix)
        print(matrix)
        if len(matrix) != 0:              #删除值并且把值全变成空的列表
            for a in range(len(matrix)): 
                matrix[a] = []
            while len(matrix) > len(matrix_1[0]):
                matrix.pop()
            while len (matrix) < len(matrix_1[0]):
                matrix.append([])
            for b in range(len(matrix_1[0])):  #填值
                for c in range(len(matrix_1)):
                    if time != len(matrix_1):
                        time += 1
                        matrix[conter].append(matrix_1[c][b])  
                    elif time == len(matrix_1):
                        time = 1
                        conter += 1
                        matrix[conter].append(matrix_1[c][b])

去除J里面的值，然后再S中查找值的个数用COUNT，求和

class Solution:
    def numJewelsInStones(self, J: str, S: str) -> int:
        num = 0
        for i in J:
            num_1 = S.count(i)
            num += num_1
        return num