2021-05-11

一、英语

1、no sooner than:一……就
2、render:致使、造成

二、算法题

1、48：坐标转换问题，数组

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        """
        #题解的过程见ipad
        #方法一：直接根据旋转得到坐标的变化
        #python中//表示整数除法，向下取整
        n = len(matrix)
        row = int(n/2)
        col = int((n+1)/2)
        for i in range(row):
            for j in range (col):
                tmp = matrix[i][j]
                matrix[i][j] = matrix[n-1-j][i]
                matrix[n-1-j][i] = matrix[n-1-i][n-1-j]
                matrix[n-1-i][n-1-j] = matrix[j][n-1-i]
                matrix[j][n-1-i] = tmp
        """
        #方法二：通过翻转实现旋转90度
        #首先上下翻转
        n = len(matrix)
        row = int(n/2)
        for i in range(row):
            for j in range(n):
                matrix[i][j],matrix[n-1-i][j] = matrix[n-1-i][j],matrix[i][j]
        #对角线翻转
        for i in range(n):
            for j in range(i):
                matrix[i][j],matrix[j][i] = matrix[j][i],matrix[i][j]

2、240:数组+排序的性质

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        #由于每一行和每一列都是有序的，从（m-1,0)开始，如果target比它小那么就向上
        #如果大，那么就向右
        m = len(matrix)
        n = len(matrix[0])
        cur = matrix[m-1][0]
        i = m-1
        j = 0
        while i>=0 and j<=(n-1):
            if cur == target:
                return True
            if cur > target:
                i-=1
                if i >= 0:
                    cur = matrix[i][j]
                else:
                    return False
            else:
                j += 1
                if j <= n-1:
                    cur = matrix[i][j]
                else:
                    return False
        return False

明天也要好好加油！
zzh生日快乐！