leetcode

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problem 26,27,28,29

26.Remove Duplicates from Sorted Array

easy

Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.

Example 1:

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Given 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 returned length.

Example 2:

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Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

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// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

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class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        if len(nums)==0:return 0
        i=1;j=1
        while i<len(nums):
            if nums[i]!=nums[i-1]:
                nums[j]=nums[i]
                j+=1
            i+=1
        nums=nums[:j]
        # print(nums)
        return j

27. Remove Element

easy

Given an array nums and a value val, 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 by modifying the input array in-place with O(1) extra memory.

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

Example 1:

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

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

It doesn't matter what you leave beyond the returned length.

Example 2:

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Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

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// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

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

28. Implement strStr()

easy

Implement strStr().

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

Example 1:

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Input: haystack = "hello", needle = "ll"
Output: 2

Example 2:

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Input: haystack = "aaaaa", needle = "bba"
Output: -1

Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C’s strstr() and Java’s indexOf().

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class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        
        if len(needle)==0:return 0
        i=0
        n=len(needle)
        while i < len(haystack):
            if len(haystack[i:])<n: return -1
            if haystack[i]==needle[0]:
                if haystack[i:i+n]==needle:
                    return i
            i+=1
        return -1

29. Divide Two Integers

meddian

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:

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

Example 2:

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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.

这个题目看上去很简单，但是如果用加减来实现除法，开销就会非常大。

但是如果把这个问题看成一个除法公式的实现，就要快很多

但是题目中又要求不用乘除，这个时候就需要用二进制的位移来代替乘除

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class Solution:
    def divide(self, dividend, divisor):
        i, a, b = 0, abs(dividend), abs(divisor)
        if a == 0 or a < b:
            return 0

        while b <= a:
            b = b << 1
            i = i + 1
        else:
            res = (1 << (i - 1)) + self.divide(a - (b >> 1), abs(divisor))
            if (dividend ^ divisor) < 0:
                res = -res
            return min(res, (1 << 31) - 1)