数组按出现频次从大到小排序
http://www.geeksforgeeks.org/amazon-interview-set-21/
#include <stdio.h>
typedef struct
{
int id;
int count;
}item;
void sort(int a[], int n)
{
item temp[n];
memset(temp,0,sizeof(temp));
int index=0;
for(int i=0;i<n;i++)
{
int count=1;
temp[index].id=a[i];
while(temp[index].id==a[i+count])
{
count++;
}
temp[index].count=count;
printf("id:%d,count:%d\n",temp[index].id,temp[index].count);
index++;
i+=count-1;
}
int j=0;
int k=0;
while(temp[j].count!=0) //should usd stable sort
{
int k=j+1;
while(temp[k].count!=0)
{
if(temp[k].count>temp[j].count)
{
item var=temp[j];
temp[j]=temp[k];
temp[k]=var;
}
k+=1;
}
printf("id:%d,count:%d\n",temp[j].id,temp[j].count);
j+=1;
}
printf("ddddd\n");
int m=0;
while(temp[m].count!=0)
{
while(temp[m].count--)
{
printf("%d",temp[m].id);
}
m+=1;
}
}
int main(void) {
// your code goes here
int a[]={65,65,1,1,1,2,2,0,9,9,9,9};
sort(a,12);
return 0;
}
344. Reverse String
Write a function that takes a string as input and returns the string reversed.
Example:
Given s = “hello”, return “olleh”.
void inverse(int start, int end, char *p)
{
char temp;
if(p)
{
temp=p[start];
p[start]=p[end];
p[end]=temp;
return;
}
else
{
return;
}
}
char* reverseString(char* s)
{
int length=strlen(s);
printf("%d\n",length);
if(!s)
{
return NULL;
}
int start;
int end;
start=0;
end=length-1;
while(start<end)
{
//char *p=s+start;
inverse(start,end,s); //这边犯了个错误，将 p作为参数传入，p地址偏移后，end的计数就应该也要减1，正确应该直接将S传入
//printf("%c",s[start]);
start++;
end--;
//printf("%s\n",s);
}
return s;
}
136. Single Number
Given an array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
int singleNumber(int* nums, int numsSize)
{
//a^a=0,a^b=b^a, 0^a=a;
//利用异或操作的特性 int a=0;
for(int i=0;i<numsSize;i++)
{
a=a^nums[i];
}
return a;
}
389. Find the Difference
Given two strings s and t which consist of only lowercase letters.
String t is generated by random shuffling string s and then add one more letter at a random position.
Find the letter that was added in t.
void fast_sort(int start, int end, char *array)
{
if(start>=end)
return;
int pivot=array[start];
int i=start;
int j=end;
while(i<j)
{
while(i<j && array[j]>=pivot)
{
j--;
}
array[i]=array[j];
while(i<j && array[i]<=pivot)
{
i++;
}
array[j]=array[i];
}
array[i]=pivot;
fast_sort(start,i-1, array);
fast_sort(i+1,end,array);
}
char findTheDifference(char* s, char* t)
{
int len1=strlen(s);
int len2=strlen(t);
char *comp1=(char *)malloc(sizeof(char)*strlen(t));
//char *comp2=(char *)malloc(sizeof(char)*strlen(t));
memcpy(comp1,t,len2);
printf("len t:%d,len s:%d\n",strlen(t),strlen(s));
printf("%s\n",s);
printf("%s\n",t);
fast_sort(0,len1-1,s);
fast_sort(0,len2-1,t);
printf("%s\n",s);
printf("%s\n",t);
for(int i=0;i<len1;i++)
{
if(s[i]!=t[i])
{
if(t[i]==comp1[len2-1]);
return s[i];
}
}
char ret=t[len2-1];
//free(comp1);
//free(comp2);
printf("%c",ret);
return ret;
}
Ransom note
看成了KMP 匹配算法，这边实现了下
void get_next(char *key,int *next)
{ next[0]=0;
int i=0;
int j=-1;
while(i<strlen(key))
{
if(j==-1 || key[i]==key[j])
{
++i;
++j;
next[i]=j;
}
else
{
j=next[j]-1;
}
}
}
bool canConstruct(char* ransomNote, char* magazine)
{
int len=strlen(ransomNote);
int len_mag=strlen(magazine);
int *next=(int *)malloc(sizeof(int)*len);
get_next(ransomNote, next);
for(int i=0;i<len;i++)
{ printf("%d\t",next[i]); }
printf("\n len:%d,len_mag:%d\n",len,len_mag);
int i=0;
int j=0;
while(i<len_mag && j<len)
{
printf("i:%d,j:%d\n",i,j);
if(magazine[i]==ransomNote[j] )
{ i++; j++; }
else
{
if(j==0)
{ i++; }
else
{ j=next[j]; }
}
if(j==len)
{ printf("true\n"); return true; }
}
printf("false\n");
return false;
}
真正的解法：
349 Intersection of Two ArraysHash 表的解法
#define HASHSIZE 1000
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int* intersection(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize)
{
int *ret;
int hash[HASHSIZE] = { 0 };
if (!nums1 || !nums2)
return NULL;
*returnSize = 0;
ret = calloc(MAX(nums1Size, nums2Size), sizeof(*nums1));
for (int i = 0; i < nums1Size; ++i)
hash[nums1[i]] = 1;
for (int i = 0; i < nums2Size; ++i)
if (hash[nums2[i]] == 1)
hash[nums2[i]] = 2;
for (int i = 0; i < HASHSIZE; ++i)
if (hash[i] == 2)
ret[(*returnSize)++] = i;
return ret;
}`
169 Majority ElementAll
Solutions
:https://discuss.leetcode.com/topic/17446/6-suggested-solutions-in-c-with-explanations- Hash table:- Moore voting algorithm** introduce to moore voting algorithm**<http://www.cs.utexas.edu/~moore/best
ideas/mjrty/index.html>
int majorityElement(int* nums, int numsSize)
{
int count=0;
int retval=0;
for(int i=0;i<numsSize;i++)
{
if(count==0)
{
retval=nums[i];
count++;
}
else
{
if(retval==nums[i])
{
count++;
}
else
{
count--;
}
}
}
return retval;
}
Bit manipulationint majorityElement(int* nums, int numsSize)
{
int mask=0x00000001;
int retval;
for(int i=0;i<32;i++)
{
int count=0;
for(int j=0;j<numsSize;j++)
{
if(mask&nums[j])
{
count++;
}
}
if(count>numsSize/2)
{
retval|=mask;
}
mask<<=1;
}
}
350. Intersection of Two Arrays IIelegant
solution of python:
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
dict1 = dict()
for i in nums1:
if i not in dict1:
dict1[i] = 1
else:
dict1[i] += 1
ret = []
for i in nums2:
if i in dict1 and dict1[i]>0:
ret.append(i)
dict1[i] -= 1
return ret
401. Binary Watch
from collections import defaultdict
class Solution(object):
def readBinaryWatch(self, num):
"""
:type num: int
:rtype: List[str]
"""
minute=defaultdict(lambda:list())
hour=defaultdict(lambda:list())
for i in range(60):
binary='{0:b}'.format(i)
minute[binary.count('1')].append(i)
for j in range(12):
binary='{0:b}'.format(j)
hour[binary.count('1')].append(j)
ret=list()
for i in range(num+1):
for j in hour[i]:
for k in minute[num-i]:
if len(str(k))<2:
ret.append(str(j)+':0'+str(k))
else:
ret.append(str(j)+':'+str(k))
return ret
566. Reshape the Matrix
class Solution(object):
def matrixReshape(self, nums, r, c):
flat = sum(nums, [])
if len(flat) != r * c:
return nums
print [iter(flat)]
print flat
#print *([iter(flat)]*c)
tuples = zip(*([iter(flat)] * c))
return map(list, tuples)
557 Reverse Words in a String III
my solution:
class Solution(object):
def reverseWords(self, s):
"""
:type s: str
:rtype: str
"""
lista=list()
s1=""
ret=""
for x in s:
if x!=' ':
s1+=x
else:
lista.append(s1)
s1=""
lista.append(s1) // append last string
print lista
for x in lista:
x=x[::-1]
ret+=x+' '
ret=ret[:-1] //delete last space, attention, last index is -1,and not included
return ret
better solution:
use python split function:
class Solution(object):
def reverseWords(self, s):
"""
:type s: str
:rtype: str
"""
return " ".join(map(lamada x: x[ : :-1], s.split()))
// note: " ".join will insert a space into join list element
62 unique path
- By matrix:
class Solution:
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
matrix = [[0 for j in range(n)] for i in range(m)]
matrix[0][0] = 1
for i in range(m):
for j in range(n):
if j > 0:
matrix[i][j] += matrix[i][j - 1]
if i > 0:
matrix[i][j] += matrix[i - 1][j]
return matrix[-1][-1]
- By hashmap:
class Solution:
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
dic = {(0, 0): 1}
i, j = 0 , 1
while i < m:
add1 = dic[(i, j - 1)] if j > 0 else 0
add2 = dic[(i - 1, j)] if i > 0 else 0
dic[(i, j)] = add1 + add2
if j + 1 < n: j += 1
else: i, j = i + 1, 0
return dic[(m - 1, n - 1)]
63. Unique Paths II
class Solution:
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
m=len(obstacleGrid)
n=len(obstacleGrid[0])
if obstacleGrid[0][0]==1:
return 0
obstacleGrid[0][0]=2
for i in range(m):
for j in range(n):
if i>0 and obstacleGrid[i][j]!=1 :
obstacleGrid[i][j]+=obstacleGrid[i-1][j]
print "i=%d,j=%d, %d" %(i,j,obstacleGrid[i][j] )
if j>0 and obstacleGrid[i][j]!=1 :
obstacleGrid[i][j]+=obstacleGrid[i][j-1]
print "xxx i=%d,j=%d, %d" %(i,j,obstacleGrid[i][j] )
return obstacleGrid[-1][-1]/2
正确方式：
class Solution:
def uniquePathsWithObstacles(self, obstacleGrid):
""
:type obstacleGrid: List[List[int]]
:rtype: int
""
m=len(obstacleGrid)
n=len(obstacleGrid[0])
if obstacleGrid[0][0]==1:
return 0
obstacleGrid[0][0]=2
for i in range(m):
for j in range(n):
if i>0 and obstacleGrid[i][j]!=1 and obstacleGrid[i-1][j]!=1:
obstacleGrid[i][j]+=obstacleGrid[i-1][j]
if j>0 and obstacleGrid[i][j]!=1 and obstacleGrid[i][j-1]!=1:
obstacleGrid[i][j]+=obstacleGrid[i][j-1]
return obstacleGrid[-1][-1]/2
when input is [0,1][0,0] the output will be 0,
python在进行二维数组遍历的时候，并非按照逐行后逐列的方式遍历，可能是先列后行。因此结果错误。
49. Group Anagrams
Tag: HashTable, String
我的code，超时，分析，时间复杂度应该是o（n^2）,(1+n)n/2
# my solution , out of time
class Solution:
def isAnagrams(self,a,b):
dic1=dict()
dic2=dict()
for c in a:
dic1[c]=dic1.get(c,0)+1
for c in b:
dic2[c]=dic2.get(c,0)+1
return dic1==dic2
def groupAnagrams(self, strs):
"""
#:type strs: List[str]
#:rtype: List[List[str]]
"""
li=list()
ret=list()
for char in strs:
a=bool()
for i in range(len(ret)):
if self.isAnagrams(char,ret[i][0]):
ret[i].append(char)
a=True
if a==False:
li.append(char)
ret.append(li)
li=[]
a=False
return ret
"""
for i in len(strs):
j=i+1
while j<len(strs):
if isAnagrams(strs[i],strs[j]):
li.add(strs[j])
if li!=[]:
li.add(str[i])
ret.add(li)
li=[]
return ret
"""
""""
从discuss中看到的，通用的做法是对元素先进行排序，然后通过hashmap进行索引，此种方法的复杂度与排序算法相关，python和C++ 都提供了O（NlogN）的复杂度的排序函数。
class Solution:
def convert(self,s):
lis=list(s)
lis.sort()
print(lis)
return ''.join(lis)
def groupAnagrams(self, strs):
dic={}
for str in strs:
cons=self.convert(str)
if cons in dic:
dic[cons].append(str)
else:
dic[cons]=[str]
return list(dic.values())
# solution 2
class Solution:
def groupAnagrams(self, strs):
"""
:type strs: List[str]
:rtype: List[List[str]]
"""
groups = {}
for s in strs:
key = ''.join(sorted([c for c in s]))
if key in groups:
groups[key].append(s)
else:
groups[key] = [s]
return list(groups.values())
# solution 3
class Solution:
def groupAnagrams(self, strs):
dic = collections.defaultdict(list)
for s in strs: dic["".join(sorted(s))].append(s)
return list(dic.values())
python3 速度最快的解决方案
class Solution:
def groupAnagrams(self, strs):
"""
:type strs: List[str]
:rtype: List[List[str]]
"""
d = {}
for s in strs:
ss = ''.join(sorted(s))
if ss not in d:
d[ss] = [s]
else:
d[ss].append(s)
ans = []
for key in d:
ans.append(d[key])
return ans
367. Valid Perfect Square
class Solution:
def isPerfectSquare(self, num):
"""
:type num: int
:rtype: bool
"""
left=0
right=num
while left<right:
mid=left+(right-left)//2
if mid **2==num:
return True
elif mid**2<num:
left=mid+1
else:
right=mid
if left==right and left **2==num:
return True
return False
solution from discussion
Solving with Bitwise trick.
def BitwiseTrick(self, num):
root = 0
bit = 1 << 15
while bit > 0 :
root |= bit
if root > num // root:
root ^= bit
bit >>= 1
return root * root == num
Using Newton's Method
def NewtonMethod(self, num):
r = num
while r*r > num:
r = (r + num/r) // 2
return r*r == num
Math Trick for Square number is 1+3+5+ ... +(2n-1)
def Math(self, num):
i = 1
while (num>0):
num -= i
i += 2
return num == 0
Binary Search Method
def BinarySearch(self, num):
left = 0
right = num
while left <= right:
mid = left + (right-left)//2
if mid ** 2 == num:
return True
elif mid ** 2 > num:
right = mid -1
else:
left = mid +1
return False
Linear Method (Naive) - For comparison
def Linear(self, num):
i = 1
while i ** 2 <= num:
if i ** 2 == num:
return True
else:
i += 1
return False
I test these five methods with for i in range(100000): function(i), and get results as below.
Time for Bitwise's Method : 0.45249104499816895
Time for Newton's Method : 0.3492701053619385
Time for Math's Method : 2.641957998275757
Time for Binary Search's Method : 1.5031492710113525
Time for Linear's Method : 17.613927125930786
6. ZigZag Conversionf
for easy understanding
/*n=numRows
Δ=2n-2 1 2n-1 4n-3
Δ= 2 2n-2 2n 4n-4 4n-2
Δ= 3 2n-3 2n+1 4n-5 .
Δ= . . . . .
Δ= . n+2 . 3n .
Δ= n-1 n+1 3n-3 3n-1 5n-5
Δ=2n-2 n 3n-2 5n-4
*/
class Solution {
public:
string convert(string s, int numRows) {
string result="";
if(numRows==1)
return s;
int step1,step2;
int len=s.size();
for(int i=0;i<numRows;++i){
step1=(numRows-i-1)*2;
step2=(i)*2;
int pos=i;
if(pos<len)
result+=s.at(pos);
while(1){
pos+=step1;
if(pos>=len)
break;
if(step1)
result+=s.at(pos);
pos+=step2;
if(pos>=len)
break;
if(step2)
result+=s.at(pos);
}
}
return result;
}
};
112 path sum
my solution:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
count=0
def hasPathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
self.count=self.count+1
if root==None and sum==0:
if self.count==1:
return False
else:
return True
if root==None and sum!=0:
return False
if root!=None:
return self.hasPathSum(root.left,sum-root.val) or self.hasPathSum(root.right,sum-root.val)
final solution:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def hasPathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
if root==None:
return False
if root.val==sum and root.left==None and root.right==None:
return True
return self.hasPathSum(root.left,sum-root.val) or self.hasPathSum(root.right,sum-root.val)
84. Largest Rectangle in Histogram
class Solution(object):
def largestRectangleArea(self, heights):
"""
:type heights: List[int]
:rtype: int
"""
heights.append(0)
stack = [-1] #add -1 to stack, this is magic code for me, keep stack never be None
ans = 0
for i in xrange(len(heights)):
while heights[i] < heights[stack[-1]]:
h = heights[stack.pop()]
w = i - stack[-1] - 1 # should minus 1 cause stack has been poped one time
ans = max(ans, h * w)
stack.append(i)
heights.pop()
return ans