【python3】从leetcode 744 & 367 & 74 & 240 &33 & 69 &29 & 441讲解二分搜索

 

 

目录

二分搜索

模版

744. Find Smallest Letter Greater Than Target

367. Valid Perfect Square

74. Search a 2D Matrix

240. Search a 2D Matrix II

33. Search in Rotated Sorted Array

69. Sqrt(x)

29. Divide Two Integers 

441. Arranging Coins

153. Find Minimum in Rotated Sorted Array

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二分搜索

二分法描述了在有序集合中搜索特定值的过程，其时间复杂度是O(logn)。

 

模版

1 

def binarySearch(nums, target):
    """
    :type nums: List[int]
    :type target: int
    :rtype: int
    """
    if len(nums) == 0:
        return -1

    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1

    # End Condition: left > right
    return -1

2 

def binarySearch(nums, target):
    """
    :type nums: List[int]
    :type target: int
    :rtype: int
    """
    if len(nums) == 0:
        return -1

    left, right = 0, len(nums)
    while left < right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid

    # Post-processing:
    # End Condition: left == right
    if left != len(nums) and nums[left] == target:
        return left
    return -1

 

---

744. Find Smallest Letter Greater Than Target

Given a list of sorted characters letters containing only lowercase letters, and given a target letter target, find the smallest element in the list that is larger than the given target.

Letters also wrap around. For example, if the target is target = 'z' and letters = ['a', 'b'], the answer is 'a'.

Examples:

Input:
letters = ["c", "f", "j"]
target = "a"
Output: "c"

Input:
letters = ["c", "f", "j"]
target = "c"
Output: "f"

Input:
letters = ["c", "f", "j"]
target = "d"
Output: "f"

Input:
letters = ["c", "f", "j"]
target = "g"
Output: "j"

Input:
letters = ["c", "f", "j"]
target = "j"
Output: "c"

Input:
letters = ["c", "f", "j"]
target = "k"
Output: "c"

class Solution:
    def nextGreatestLetter(self, letters, target):
        """
        :type letters: List[str]
        :type target: str
        :rtype: str
        """
        #find the insert position
        lo, hi = 0,len(letters)-1
        while(lo<=hi):
            mid = int((lo+hi)/2)
            if letters[mid] > target:hi=mid-1
            else:lo = mid+1
        
        return letters[lo%len(letters)]

---

367. Valid Perfect Square

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

Input: 16
Output: true

Example 2:

Input: 14
Output: false

class Solution:
    def isPerfectSquare(self, num):
        """
        :type num: int
        :rtype: bool
        """
        lo = 0;hi = num
        while(lo <= hi):
            mid = int((lo+hi)/2)
            if mid**2 == num:return True
            elif mid**2 > num:hi = mid -1
            else:lo = mid+1
        return False
    

74. Search a 2D Matrix

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

双重binary search

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or not matrix[0]:return False
        row_lo = 0;row_hi = len(matrix)-1
        col_lo = 0;col_hi = len(matrix[0])-1
        while(row_lo<=row_hi):
            mid = (row_lo + row_hi) // 2
            if matrix[mid][0] == target:return True
            elif matrix[mid][0] > target:row_hi = mid-1
            else:row_lo = mid+1
                
        #lower or larger 
        if row_lo-1 < 0:return False
        while(col_lo <= col_hi):
            mid  = (col_lo + col_hi) // 2
            if matrix[row_lo-1][mid] == target:return True
            elif matrix[row_lo-1][mid]>target:col_hi = mid-1
            else:col_lo = mid+1
        return False
        

240. Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

遍历每行再binary search，slow 

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or not matrix[0]:return False
        for row in range(len(matrix)):
            col_lo = 0;col_hi = len(matrix[0])-1
            while(col_lo <= col_hi):
                mid  = (col_lo + col_hi) // 2
                if matrix[row][mid] == target:return True
                elif matrix[row][mid]>target:col_hi -= 1
                else:col_lo += 1
        return False

33. Search in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

class Solution:
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        if target not in nums:return -1
        minpos = nums.index(min(nums))
        part1 = nums[:minpos]
        part2 = nums[minpos:]
        
        if target in part1:
            lo = 0;hi = minpos-1
            while(lo<=hi):
                mid = (lo+hi) // 2
                if nums[mid] == target:return mid
                elif nums[mid] > target:hi = mid-1
                else:lo = mid+1
        else:
            lo = 0;hi=len(part2)-1
            while(lo <= hi):
                mid = (lo+hi) // 2
                if part2[mid] == target:return mid+minpos
                elif part2[mid] > target:hi = mid-1
                else:lo = mid+1

69. Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

class Solution:
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        if x==1 or x == 0:return x
        lo = 1;hi = x
        while(lo<=hi):
            mid = (lo+hi) // 2
            if mid**2 <= x and (mid+1)**2 > x:return mid
            elif mid**2 > x:hi = mid -1
            else:lo = mid+1

29. Divide Two Integers 

Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.

Return the quotient after dividing dividend by divisor.

The integer division should truncate toward zero.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3

Example 2:

Input: dividend = 7, divisor = -3
Output: -2

Note:

• Both dividend and divisor will be 32-bit signed integers.
• The divisor will never be 0.
• Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 231 − 1 when the division result overflows.

class Solution:
    def divide(self, dividend, divisor):
        """
        :type dividend: int
        :type divisor: int
        :rtype: int
        """
        if dividend == 0:return 0
        tag = -1 if dividend*divisor < 0 else 1
        dividend, divisor = abs(dividend),abs(divisor)
        MAX = (1<<31)-1
        MIN = -1*(1<<31)
        lo = 1;hi = dividend
        while(lo <= hi):
            mid = (lo+hi) // 2
            if mid*divisor <=dividend and (mid+1) * divisor>dividend:
                if (tag == -1 and mid*tag>=MIN) or (tag==1 and mid*tag<=MAX):return mid*tag
                else: return MAX
            elif mid*divisor > dividend:hi = mid-1
            else:lo = mid+1
        return 0
        

441. Arranging Coins

You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.

Given n, find the total number of full staircase rows that can be formed.

n is a non-negative integer and fits within the range of a 32-bit signed integer.

Example 1:

n = 5

The coins can form the following rows:
¤
¤ ¤
¤ ¤

Because the 3rd row is incomplete, we return 2.

 

Example 2:

n = 8

The coins can form the following rows:
¤
¤ ¤
¤ ¤ ¤
¤ ¤

Because the 4th row is incomplete, we return 3.

class Solution(object):
    def arrangeCoins(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 0 or n == 1:return n

        lo = 1;hi = n
        while(lo <= hi):
            mid = (lo + hi) // 2
            totalcoins = mid*(mid+1)/2 
            leave = n - totalcoins
            if totalcoins <= n and leave < (mid+1):return mid
            elif totalcoins > n:hi = mid -1
            else:lo = mid + 1

 

153. Find Minimum in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e.,  [0,1,2,4,5,6,7] might become  [4,5,6,7,0,1,2]).

Find the minimum element.

You may assume no duplicate exists in the array.

Example 1:

Input: [3,4,5,1,2] 
Output: 1

Example 2:

Input: [4,5,6,7,0,1,2]
Output: 0

# The isBadVersion API is already defined for you.
# @param version, an integer
# @return a bool
# def isBadVersion(version):

class Solution:
    def firstBadVersion(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n==0:return 0
        lo = 1;hi = n
        while(lo <= hi):
            mid = (lo + hi) // 2
            if isBadVersion(mid) == True:hi = mid -1
            else:lo = mid + 1
        return hi+1