劍指offer其餘題目總結1....先補上二叉搜索樹的第k個結點

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    int kth;
    TreeNode * ans;
    void dfs(TreeNode * p){
        if(p==NULL||kth<1)return;
        dfs(p->left);
        if(kth==1)ans=p;
        if(--kth>0)dfs(p->right);
    }
    TreeNode* KthNode(TreeNode* pRoot, int k)
    {
        if(pRoot==NULL)return NULL;
        ans=NULL;
        kth=k;
        dfs(pRoot);
        return ans;
    }

    
};

再附上非遞歸的寫法..可以更理解這個過程

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    
    TreeNode* KthNode(TreeNode* pRoot, int k)
    {
        if(pRoot==NULL)return NULL;
        stack<TreeNode *>sta;
        TreeNode * p=pRoot;
        while(!sta.empty()||p)
        {
            while(p)
            {
                sta.push(p);
                p=p->left;
            }
            TreeNode * top=sta.top();sta.pop();
            if((--k)==0)return top;
            p=top->right;
        }
        return NULL;
    }

    
};

（1）二維數組中的查找

class Solution {
public:
    
    bool Find(int target, vector<vector<int> > array) {
        int leny=array[0].size()-1;
        int lenx=array.size()-1;
        if(lenx==-1||leny==-1)return false;
        int y=0,x=lenx;
        while(x>=0&&y<=leny&&array[x][y]!=target)
        {
            if(array[x][y]<target)y++;
            else x--;
             
        }
        if(x>=0&&y<=leny)return true;return false;
    }
};

（2）替換空格

class Solution {
public:
	void replaceSpace(char *str,int length) {
        char ans[length*3];
        int k=0;
        for(int i=0;i<length;i++)
        {
            if(str[i]==' ')
            {
                ans[k++]='%';
                ans[k++]='2';
                ans[k++]='0';
            }
            else
            {
                ans[k++]=str[i];
            }
        }
        for(int i=0;i<k;i++)
        {
            str[i]=ans[i];
        }
        str[k]='\n';
	}
};

（3）用兩個棧實現隊列

class Solution
{
public:
    void push(int node) {
        stack1.push(node);
    }

    int pop() {
        while(!stack1.empty())
        {
            stack2.push(stack1.top());
            stack1.pop();
        }
        int a=stack2.top();
        stack2.pop();
        while(!stack2.empty())
        {
            stack1.push(stack2.top());
            stack2.pop();
        }
        return a ;
    }

private:
    stack<int> stack1;
    stack<int> stack2;
};

（4）旋轉數組最小值

class Solution {
public:
    
    int minNumberInRotateArray(vector<int> rotateArray) {
        if(rotateArray.size()==0)return 0;
        int ans=rotateArray[0];
        int l=0,r=rotateArray.size()-1;
        while(r>l)
        {
            int mid=(r+l)/2;
            if(rotateArray[mid]>rotateArray[l]){l=mid;continue;}
            if(rotateArray[mid]<rotateArray[r]){r=mid;continue;}
            if(rotateArray[l]>rotateArray[r])break;
        }
        ans=min(ans,rotateArray[r]);
        return ans;
    }
};

（5）斐波那契數列

class Solution {
    
public:
    int Fibonacci(int n) {
        if(n==0)return 0;
        if(n<=2)return 1;
        return Fibonacci(n-1)+Fibonacci(n-2);
    }
};

480ms 771k

class Solution {
    
public:
    int Fibonacci(int n) {
        if(n==0)return 0;
        if(n<=2)return 1;
        int a=1,b=1;
        n-=2;
        while(n--)
        {
            a=a+b;
            b=a-b;
        }
        return a;
    }
};

5ms  480

不僅避免了遞歸調用的棧消耗而且時間消耗也短了很多...仔細想想的話上面的遞歸是很蠢的.相當於每一個都從n回推到0和1，真的很蠢.我很抱歉.....

（6）跳臺階

class Solution {
public:
    int jumpFloor(int number) {
        if(number==1)return 1;
        if(number==2)return 2;
        return jumpFloor(number-1)+jumpFloor(number-2);
    }
};

608 488k....這裏....改成非遞歸試試.

class Solution {
public:
    int jumpFloor(int number) {
        if(number==1)return 1;
        if(number==2)return 2;
        number-=2;
        int a=2,b=1;
        while(number--)
        {
            a=a+b;
            b=a-b;
        }
        return a;
    }
};

4 492

（7）變態跳臺階

class Solution {
public:
    int jumpFloorII(int number) {
        return 1<<(number-1);
    }
};

（8）矩形覆蓋

class Solution {
public:
    int rectCover(int number) {
        int n=number;
         int a=2;
        int b=1;
        if(n<=0)return 0;
        if(n==1)return 1;
        if(n==2)return 2;
        else
            
        {
            n-=2;
            while(n--)
            {
                b=a+b;
                swap(a,b);
            }
            return a;
        }
    
    }
};

（9）二進制中1的個數

class Solution {
public:
     int  NumberOf1(int n) {
         int ans=0;
         for(int i=0;i<=31;i++)
         {
             if((1<<i)&n)ans++;
         }
         return ans;
     }
};

(8)實現pow

class Solution {
public:
    double Power(double base, int exponent) {
    if(base==0)return 0;
    if(exponent==0)return 1;
    if(exponent<0){exponent*=-1;base=1.0/base;}
        double ans=1.0;
    while(exponent)
    {
        if(exponent%2==1){ans*=base;exponent--;}
        if(exponent!=0){
            exponent/=2;
        base=base*base;
        }
        
    }
        return ans;
    }
};

不想寫了  我要吐了  剩下的再說吧。。。。反正暑期實習沒機會了  4.20前報名  我人傻了  明明6月底纔去，，這麼早，，看來就業形勢嚴峻啊。。。。