Cambridge offer series (1~10)

1. Title description in a two-dimensional array, each row is ordered in ascending order from left to right, and each column is sorted in ascending order from top to bottom. Complete a function, enter a two-dimensional array and an integer to determine if the array contains the integer. Idea: From the bottom of the left, small, upward, large, to the right.

classSolution { Public:    BOOLFind (vector<vector<int> > Array,inttarget) {                if(array.size () = =0)return false; intR=array.size (); intc=array[0].size (); inti=r-1, j=0;  while(i<r&&i>-1&&j<c&&j>-1){                        if(Array[i][j] = = target)return true; if(array[i][j]>target) {i--; }Else{J++; }        }        return false; }};

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2. Title Description Please implement a function to replace the space in a string with "%20". For example, when the string is we are Happy. The string after substitution is we%20are%20happy. Idea: Replace spaces.

//Length is the maximum length for a cow system to specify a string output, fixed to a constantclassSolution { Public:    voidReplacespace (Char*STR,intlength) {    inti =0; intj =0; intNspace =0; Char*pstr =NULL; PStr= (Char*)malloc(sizeof(Char) *length *3);  for(i =0, j =0; i<length; i++, J + +)    {        if(' '==Str[i]) {Pstr[j]='\%'; pstr[++J] ='2'; pstr[++J] ='0'; }        Else{Pstr[j]=Str[i]; }    }     for(i=0; i<j;i++) {Str[i]=Pstr[i]; }     Free(PSTR); PStr=NULL;}};

View Code3. Title Description Enter a list of values for each node of the list to print from the end of the header. Idea: Use vector to pick up, then call reverse algorithm reverse (v.begin (), V.end ());

/** * struct ListNode {* int val;* struct ListNode *next;* listnode (int x):* val (x), Next (NULL) {*}*};*/classSolution { Public: Vector<int> Printlistfromtailtohead (structlistnode*head) {                structlistnode* tmp =Head; Vector<int>v;  while(TMP! =NULL) {V.push_back (TMP-val); TMP= tmp->Next;        } reverse (V.begin (), V.end ()); returnv; }};

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4. The topic describes the input of a binary tree of the pre-sequence traversal and the results of the middle sequence traversal, please reconstruct the two fork tree. Assume that no duplicate numbers are included in the result of the input's pre-order traversal and the middle-order traversal.  For example, enter the pre-sequence traversal sequence {1,2,4,7,3,5,6,8} and the middle sequence traversal sequence {4,7,2,1,5,3,8,6}, then rebuild the binary tree and return. Idea: Through the preamble and the middle order, find the root, and then recursive call.

/** Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * T Reenode (int x): Val (x), left (null), right (NULL) {} *}; */classSolution { Public:    structtreenode* Reconstructbinarytree (vector<int> pre,vector<int>inch) {        intIn_size =inch. Size (); if(In_size = =0)            returnNULL; Vector<int>Pre_left, Pre_right, In_left, in_right; intval = pre[0]; TreeNode* node =NewTreeNode (Val);//root node is the first element in pre        intp =0;  for(P < in_size; + +)p) {            if(inch[P] = = val)//Find The root position in                 Break; }         for(inti =0; i < in_size; ++i) {            if(I <p) {In_left.push_back (inch[i]);//Construct The left pre and inPre_left.push_back (Pre[i +1]); }            Else if(I >p) {In_right.push_back (inch[i]);//Construct The right pre and inPre_right.push_back (Pre[i]); }} node->left =Reconstructbinarytree (Pre_left, in_left); Node->right =Reconstructbinarytree (Pre_right, in_right); returnnode; }  };

View Code5. Title description use two stacks to implement a queue, complete the queue push and pop operations. The elements in the queue are of type int. Idea: The data passes in and out, after 2 stacks can be implemented in a queue.

classsolution{ Public:    voidPushintnode)    {Stack1.push (node); }    intpop () {if(Stack2.empty ()) { while(!Stack1.empty ())                {Stack2.push (Stack1.top ());            Stack1.pop (); }        }        intnode =Stack2.top ();        Stack2.pop (); returnnode; }Private: Stack<int>Stack1; Stack<int>Stack2;};

View Code6. The title describes moving a number of elements at the beginning of an array to the end of the array, which we call the rotation of the array.
Enter a rotation of a non-descending sorted array, outputting the smallest element of the rotated array.
For example, the array {3,4,5,1,2} is a rotation of {1,2,3,4,5}, and the minimum value of the array is 1.
Note: All elements given are greater than 0, and if the array size is 0, return 0. Idea: To find the mutation point, the next is the desired value.

classSolution { Public:    intMinnumberinrotatearray (vector<int>Rotatearray) {        intI=0; intLen =rotatearray.size (); if(len==0)return 0; if(len==1)returnrotatearray[0];  while(i<len-1){            if(rotatearray[i]<=rotatearray[i+1]) {i++; }Else{                 Break; }        }       returnrotatearray[i+1]; }};

View Code7. Title Description Everyone knows the Fibonacci sequence and now asks for an integer n, please output the nth of the Fibonacci sequence. N<=39 Fibonacci Series, also known as the Golden Section, refers to a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 、...... Mathematically, the Fibonacci sequence is defined as a recursive method: F (0) =0,f (1) =1,f (n) =f (n-1) +f (n-2) (n≥2,n∈n*)

classSolution { Public:    intFibonacci (intN) {if(n = =0)            return 0; if(n = =1)            return 1; intNUMFN1 =0, numfn2 =1; intCurrentnum;  for(intI=2; i<=n; ++i) {currentnum= numfn1+numfn2; Numfn1=numfn2; Numfn2=Currentnum; }        returnCurrentnum; }};

View Code8. The title describes a frog can jump up to 1 steps at a time, can also jump on the 2 level. Ask the frog to jump on an n-level step with a total number of hops. Thinking: With the 7th question

classSolution { Public:    intJumpfloor (intNumber ) {        if(Number = =1)return 1; if(Number = =2)return 2; intVal; intnum1=1; intNum2=2;  while(number-->2) {Val= num1+num2; NUM1=num2; Num2=Val; }        returnVal; }};

View Code9. The title describes a frog can jump up to 1 steps at a time, can also jump on the 2 level ... It can also jump on n levels. Ask the frog to jump on an n-level step with a total number of hops. Idea: a[n]=a[n-1]+a[n-2]+......+a[1]; ....... ............ ①a[n-1]= a[n-2]+......+a[1], ......... .......... ② two-type subtraction: a[n]=2*a[n-1];

class Solution {public:    int jumpfloorii (int number ) {          int f=1, fn=1;          for (int i=2; i<=number;i++) {            fn=2*F;            F=fn;        }         return fn;    }};

View Code10. The title describes that we can use a small rectangle of 2*1 to cover the larger rectangle horizontally or vertically. What is the total number of ways to cover a large rectangle of 2*n with n 2*1 small rectangles without overlapping? Idea: the same 7

classSolution { Public:    intRectcover (intNumber ) {        if(number==1)return 1; if(number==2)return 2; intn1=1, n2=2, val=0;  for(intI=2; i<number;i++) {Val= n1 +N2; N1=N2; N2=Val; }        returnVal; }};

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Cambridge offer series (1~10)