Classical algorithm interview questions 1-3

 

 

# 1 is inscribed: determining whether two strings is recombined same letter O (N2)
 # 1 solution, directly search an element elements 1 and Comparative
DEF View (S, T):
    S = List (S)
    T = List (T)
    R & lt True =
    R1 = False
    IF len (S) == len (T):
        for I in Range (len (S)):
            for J in Range (len (T)):
                IF S [I] == T [J]:
                   t.remove (T [J])
                   R1 = True
                   BREAK
                the else:
                    R1 = False
            R & lt R1 = R & lt &
            IF R & lt:
                Continue
            the else:
                BREAK
    the else:
        R & lt = False
    return R & lt
print(view("anangem","naangme"))

#解决方法2,使用排序的思想,直接对字符串进行排序操作即可
def  valid_anagram(s,t):
    ss=list(s)
    tt=list(t)
    ss.sort()
    tt.sort()
    return ss==tt
print(valid_anagram("anangem","naangme"))

#解决方法3，使用sorted函数，直接返回一个新列表O（nlogn）
def v3(s,t):
    return sorted(list(s))==sorted(list(t))
print(v3("anangem","naangme"))

#解决方法4，统计字母数目,技计数排序O（n）
def v4(s,t):
    d1={}
    d2={}
    for ch in s:
        d1[ch]=d1.get(ch,0)+1   #开始统计数目进行比较
    for ch in t:
        d2[ch]=d2.get(ch,0)+1
    return d1==d2
print(v4("anangem","naangme"))

#面试题目2，直接在二维列表中一目标元素
#解决方法1：直接一个一个查找摸一个元素,O(mn)
def search(matrx,target):
    for line in matrx:
        if target in line:
            return True
    return False
import numpy
x=[[1,2,3],[4,5,6]]
print(x)
print(search(x,100))

#解决方法2：先对列表进行排序，再用二分查找
def search2(matrx,target):
    h=len(matrx)
    if h==0:
        return False
    w=len(matrx[0])
    if w==0:
        return False
    left=0
    right=h*w-1
    while left<=right:
        mid = (left + right) // 2
        row = mid // w
        column = mid % w
        if matrx[row][column]==target:
            return True
        elif matrx[row][column]<target:
            left=mid+1
        elif matrx[row][column]>target:
            right=mid-1
    else:
        return False
print(x)
print(search2(x,8))

#面试题3：在列表中查找两个数的和为某一输入数的两个数索引
#解决办法1：使用循环进行搜索
def cha(nums,target):
    n=len(nums)
    for i in range(n):
        num1=target-nums[i]
        m=nums[i]
        nums.remove(nums[i])
        if num1 in nums:
            return (i,1+nums.index(num1))
        else:
            nums.insert(i,m)
print(cha([3,2,4],6))

#解决办法2：使用双层循环进行搜索
def cha2(nums,target):
    n=len(nums)
    for i in range(n):
        for j in range(i):
            if nums[i]+nums[j]==target:
                return sorted([i,j])