26-30LeetCode：Python解题

Posted 2023-02-25 Joe-Han

26. Remove Duplicates from Sorted Array【Easy】

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn’t matter what you leave beyond the new length.

Solution:

class Solution:
    # @param a list of integers
    # @return an integer
    def removeDuplicates(self, A):
        if not A:
            return 0

        index = 0

        for i in range(1, len(A)):
            if A[i] != A[newTail]:
                newTail += 1
                A[newTail] = A[i]

        return newTail + 1

27. Remove Element【Easy】

Given an array and a value, remove all instances of that value in place and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Example:

Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

Solution :

class Solution(object):
  def removeElement(self, nums, val):
    """
    :type nums: List[int]
    :type val: int
    :rtype: int
    """
    start, end = 0, len(nums) - 1
    while start <= end:
        if nums[start] == val:
            nums[start], nums[end], end = nums[end], nums[start], end - 1
        else:
            start +=1
    return start

28. Implement strStr()【Easy】

Implement strStr().

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Solution :

class Solution(object):
    def strStr(self, haystack, needle):
        """
        :type haystack: str
        :type needle: str
        :rtype: int
        """
        for i in range(len(haystack) - len(needle)+1):
            if haystack[i:i+len(needle)] == needle:
                return i
        return -1

29. Divide Two Integers【Medium】

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return MAX_INT.

Solution :

class Solution:
# @return an integer
def divide(self, dividend, divisor):
    positive = (dividend < 0) is (divisor < 0)
    dividend, divisor = abs(dividend), abs(divisor)
    res = 0
    while dividend >= divisor:
        temp, i = divisor, 1
        while dividend >= temp:
            dividend -= temp
            res += i
            i <<= 1
            temp <<= 1
    if not positive:
        res = -res
    return min(max(-2147483648, res), 2147483647)

30. Substring with Concatenation of All Words【Hard】

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given:
s: “barfoothefoobarman”
words: [“foo”, “bar”]

You should return the indices: [0,9](order does not matter).

Solution :

class Solution(object):
    def findSubstring(self, s, words):
        """
        :type s: str
        :type words: List[str]
        :rtype: List[int]
        """
        if not s or words==[]:
            return []
        lenstr=len(s)
        lenword=len(words[0])
        lensubstr=len(words)*lenword
        times=
        for word in words:
            if word in times:
                times[word]+=1
            else:
                times[word]=1
        ans=[]
        for i in xrange(min(lenword,lenstr-lensubstr+1)):
            self.findAnswer(i,lenstr,lenword,lensubstr,s,times,ans)
        return ans
    def findAnswer(self,strstart,lenstr,lenword,lensubstr,s,times,ans):
        wordstart=strstart
        curr=
        while strstart+lensubstr<=lenstr:
            word=s[wordstart:wordstart+lenword]
            wordstart+=lenword
            if word not in times:
                strstart=wordstart
                curr.clear()
            else:
                if word in curr:
                    curr[word]+=1
                else:
                    curr[word]=1
                if curr[word]>times[word]:
                    curr[s[strstart:strstart+lenword]]-=1
                    strstart+=lenword
                if wordstart-strstart==lensubstr:
                    ans.append(strstart)

Discussion:

首先确定子字符串可能的起点位置i，然后从每个i出发寻找包含words中所有单词的子字符串区间。