LeetCode---Binary Gap、Self Dividing Numbers、Reach a Number、Rectangle Overlap

868. Binary Gap

Given a positive integer N, find and return the longest distance between two consecutive 1's in the binary representation of N.

If there aren't two consecutive 1's, return 0.

Example 1:

Input: 22
Output: 2
Explanation: 
22 in binary is 0b10110.
In the binary representation of 22, there are three ones, and two consecutive pairs of 1's.
The first consecutive pair of 1's have distance 2.
The second consecutive pair of 1's have distance 1.
The answer is the largest of these two distances, which is 2.

Example 2:

Input: 5
Output: 2
Explanation: 
5 in binary is 0b101.

Example 3:

Input: 6
Output: 1
Explanation: 
6 in binary is 0b110.

Example 4:

Input: 8
Output: 0
Explanation: 
8 in binary is 0b1000.
There aren't any consecutive pairs of 1's in the binary representation of 8, so we return 0.

给定一个正整数 N，找到并返回 N 的二进制表示中两个连续的 1 之间的最长距离。 

如果没有两个连续的 1，返回 0 。 

class Solution(object):
    def binaryGap(self, N):
        """
        :type N: int
        :rtype: int
        """
        bn=bin(N)[2:]
        n=len(bn)
        ans=0
        last=-1
        for i in range(n):
            if bn[i]=='1':
                if last==-1:
                    last=i
                else:
                    if ans<=i-last:
                        ans=i-last
                    last=i
        return ans 

 728. Self Dividing Numbers

A self-dividing number is a number that is divisible by every digit it contains.

For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.

Also, a self-dividing number is not allowed to contain the digit zero.

Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.

Example 1:

Input: 
left = 1, right = 22
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]

自除数 是指可以被它包含的每一位数除尽的数。

例如，128 是一个自除数，因为 128 % 1 == 0，128 % 2 == 0，128 % 8 == 0。

还有，自除数不允许包含 0 。

给定上边界和下边界数字，输出一个列表，列表的元素是边界（含边界）内所有的自除数。

class Solution(object):
    def selfDividingNumbers(self, left, right):
        """
        :type left: int
        :type right: int
        :rtype: List[int]
        """
        res=[]
        for i in range(left,right+1):
            strnum=str(i)
            flag=True
            for digit in strnum:
                if digit=='0':
                    flag=False
                    break
                if i%int(digit)!=0:
                    flag=False
                    break
            if flag:
                res.append(i)
        return res
        

754. Reach a Number

You are standing at position 0 on an infinite number line. There is a goal at position target.

On each move, you can either go left or right. During the n-th move (starting from 1), you take nsteps.

Return the minimum number of steps required to reach the destination.

Example 1:

Input: target = 3
Output: 2
Explanation:
On the first move we step from 0 to 1.
On the second step we step from 1 to 3.

 

Example 2:

Input: target = 2
Output: 3
Explanation:
On the first move we step from 0 to 1.
On the second move we step  from 1 to -1.
On the third move we step from -1 to 2.

参考

在一根无限长的数轴上，你站在0的位置。终点在target的位置。

每次你可以选择向左或向右移动。第 n 次移动（从 1 开始），可以走 n 步。

返回到达终点需要的最小移动次数。

class Solution(object):
    def reachNumber(self, target):
        """
        :type target: int
        :rtype: int
        """
        target=abs(target)
        k=0
        sum=0
        while sum<target:
            k+=1
            sum+=k
        d=sum-target
        if d%2==0:
            return k
        return k+1+(k%2)
        

836. Rectangle Overlap

A rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) are the coordinates of its bottom-left corner, and (x2, y2) are the coordinates of its top-right corner.

Two rectangles overlap if the area of their intersection is positive.  To be clear, two rectangles that only touch at the corner or edges do not overlap.

Given two (axis-aligned) rectangles, return whether they overlap.

Example 1:

Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
Output: true

Example 2:

Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
Output: false

矩形以列表 [x1, y1, x2, y2] 的形式表示，其中 (x1, y1) 为左下角的坐标，(x2, y2) 是右上角的坐标。

如果相交的面积为正，则称两矩形重叠。需要明确的是，只在角或边接触的两个矩形不构成重叠。

给出两个矩形，判断它们是否重叠并返回结果。

class Solution(object):
    def isRectangleOverlap(self, rec1, rec2):
        """
        :type rec1: List[int]
        :type rec2: List[int]
        :rtype: bool
        """
        d1=min(rec1[2],rec2[2])-max(rec1[0],rec2[0])
        d2=min(rec1[3],rec2[3])-max(rec1[1],rec2[1])
        if d1>0 and d2>0:
            return True
        return False