Chap6: question38 - 42

38. 数字 k 在有序数组中出现的次数

二分查找：找出第一个 k 和最后一个 k 。

#include <iostream>
using namespace std;
int getFirstOfK(int data[], int length, int k, int low, int high)
{
    if(low <= high)
    {
        int mid = (low + high) / 2;
        if(data[mid] == k && (mid == 0 || data[mid-1] != k))
            return mid;
        else if(data[mid] < k)
            low = mid + 1;
        else high = mid - 1;
        return getFirstOfK(data, length, k, low, high);
    }
    return -1;
}
int getLastOfK(int data[], int length, int k, int low, int high)
{
    if(low <= high)
    {
        int mid = (low + high) / 2;
        if(data[mid] == k && (mid == length-1 || data[mid+1] != k))
            return mid;
        else if(data[mid] > k)
            high = mid - 1;
        else low = mid + 1;
        return getLastOfK(data, length, k, low, high);
    }
    return -1;
}
int getNumberOfK(int data[], int length, int k)
{
    int count = 0;
    if(data != NULL && length > 0)
    {
        int first = getFirstOfK(data, length, k, 0, length-1);
        if(first == -1) return -1;

        int last = getLastOfK(data, length, k, first, length-1);
        count = last - first + 1;
    }
    return count;
}
int main()
{
    int data[] = {1, 2, 3, 3, 3, 3, 4, 5};
    cout << getNumberOfK(data, sizeof(data)/4, 3) << endl;
    cout << getNumberOfK(data, sizeof(data)/4, 1) << endl;
    cout << getNumberOfK(data, sizeof(data)/4, 2) << endl;
    cout << getNumberOfK(data, sizeof(data)/4, 5) << endl;

    return 0;
}

39. 二叉树的深度 && 平衡二叉树的判断 &&  二叉树结点的最大距离（题目来自编程之美，解法自创）

note：三种算法都必须是后序遍历。

#include <iostream>
#include <string>
using namespace std;
typedef struct BTNode
{
    int v;  // default positive Integer.
    BTNode *pLeft;
    BTNode *pRight;
    BTNode(int x) : v(x), pLeft(NULL), pRight(NULL) {}
} BinaryTree;
/********************************************************/
/*****        Basic functions          ***********/
BinaryTree* createBinaryTree() // input a preOrder traversal sequence, 0 denote empty node.
{
    BTNode *pRoot = NULL;
    int r;
    cin >> r;
    if(r != 0)         // equal to if(!r) return;
    {
        pRoot = new BTNode(r);
        pRoot->pLeft = createBinaryTree();
        pRoot->pRight = createBinaryTree();

    }
    return pRoot;
}
void release(BinaryTree *root){
    if(root == NULL) return;
    release(root->pLeft);
    release(root->pRight);
    delete[] root;
    root = NULL;
}
void print(BinaryTree *root, int level = 1){
    if(root == NULL) { cout << "NULL"; return; };
    string s;
    for(int i = 0; i < level; ++i) s += "   ";
    cout << root->v << endl << s;
    print(root->pLeft, level+1);
    cout << endl << s;
    print(root->pRight, level+1);
}
/******************************************************************/
int getDepth(BinaryTree *root) // leaf Node is at depth 1
{
    if(root == NULL) return 0;
    int leftDepth = getDepth(root->pLeft);
    int rightDepth = getDepth(root->pRight);
    return  1 + (leftDepth > rightDepth ? leftDepth : rightDepth);
}

bool isBalanced(BinaryTree *root, int *depth) // must be postOrder traversal
{
    if(root == NULL) { *depth = 0; return true; };
    int leftDepth, rightDepth;
    if(isBalanced(root->pLeft, &leftDepth) && isBalanced(root->pRight, &rightDepth))
    {
        *depth = 1 + (leftDepth > rightDepth ? leftDepth : rightDepth);
        if(leftDepth - rightDepth >= -1 && leftDepth - rightDepth <= 1)
            return true;
        else
            return false;
    }
}
bool isBalanced(BinaryTree *root)
{
    int depth;
    return isBalanced(root, &depth);
}

int getMaxDistance(BinaryTree *root, int *maxDistance) // leaf node depth is set to 0
{
    if(root == NULL) return -1;
    int leftDepth = getMaxDistance(root->pLeft, maxDistance);
    int rightDepth = getMaxDistance(root->pRight, maxDistance);
    if(*maxDistance < 2 + leftDepth + rightDepth)
        *maxDistance = 2 + leftDepth + rightDepth;
    return 1 + (leftDepth > rightDepth ? leftDepth : rightDepth);
}
int getMaxDistance(BinaryTree *root)
{
    int maxDistance = 0;
    getMaxDistance(root, &maxDistance);
    return maxDistance;
}
int main(){
    int TestTime = 3, k = 1;
    while(k <= TestTime)
    {
        cout << "Test " << k++ << ":" << endl; 

        cout << "Create a tree: " << endl;
        BinaryTree *pRoot = createBinaryTree();
        print(pRoot);
        cout << endl; 

        cout << "The depth of binary tree: " << getDepth(pRoot) << endl;

        if(isBalanced(pRoot))
            cout << "Does the tree is a balanced binary tree ? true" << endl;
        else
            cout << "Does the tree is a balanced binary tree ? false" << endl;

        cout << "The max distance between two nodes: " << getMaxDistance(pRoot) << endl; 

        release(pRoot);
    }
    return 0;
}

40. 数组中只出现一次的数字

首先， 参考 Link: Single Number

其次，数组中有两个只出现一次的数字时：  例：{2，4，3，6，3，2，5，5}

#include <iostream>
using namespace std;
void findTwoNumbers(int data[], int length, int *num1, int *num2)
{
    if(data == NULL || length < 2)  return;
    int total = 0;
    for(int i = 0; i < length; ++i)
        total ^= data[i];
    int shift1 = 1;
    for(int i = 0; i < sizeof(int)*8; ++i)
    {
        total >>= 1;
        shift1 <<= 1;
        if(total & 1)
            break;
    }
    *num1 = *num2 = 0;
    for(int i = 0; i < length; ++i)
    {
        if(data[i] & shift1) *num1 ^= data[i];
        else *num2 ^= data[i];
    }
}
int main(){
    int num1, num2;
    int test1[8] = { 2, 4, 3, 6, 3, 2, 5, 5};
    findTwoNumbers(test1, 8, &num1, &num2);
    cout << num1 << " "<< num2 << endl;
    return 0;
}

41. 和为 S 的连续正数序列。

#include <iostream>
using namespace std;
void numsSumToS(int S)
{
    int low = 1, high = 2;
    while(low < high)
    {
        int curSum = 0;
        for(int i = low; i <= high; ++i)
            curSum += i;
        if(curSum < S) ++high;
        else if(curSum > S) ++low;
        else
        {
            for(int i = low; i <= high; ++i)
                cout << i << '\t';
            cout << endl;
            ++high;
        }
    }
}
int main(){
    int S;
    while(true)
    {
        cout << "cin >> ";
        cin >> S;
        numsSumToS(S);
    }
    return 0;
}

42. 翻转单词顺序 && 字符串左旋转

note：左旋转 k 位相当于右旋转 N – k 位， N 为字符串长度。

Link:   7. Reverse Words in a String