剑指offer(26-30)编程题

• 二叉搜索树与双向链表
• 字符串的排列
• 数组中出现次数超过一半的数字
• 最小的K个数
• 连续子数组的最大和

26.输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的

 /*
 struct TreeNode {
     int val;
     struct TreeNode *left;
     struct TreeNode *right;
     TreeNode(int x) :
             val(x), left(NULL), right(NULL) {
     }
 };*/
 class Solution {
 public:
     vector<TreeNode*> doConvert(TreeNode* pRootOfTree) {
         vector<TreeNode*> res();
         if (pRootOfTree->left == nullptr) {
             res[] = pRootOfTree;
         } else {
             vector<TreeNode*> leftRes = doConvert(pRootOfTree->left);
             res[] = leftRes[];
             leftRes[]->right = pRootOfTree;
             pRootOfTree->left = leftRes[];
         }

         if (pRootOfTree->right == nullptr) {
             res[] = pRootOfTree;
         } else {
             vector<TreeNode*> rightRes = doConvert(pRootOfTree->right);
             res[] = rightRes[];
             pRootOfTree->right = rightRes[];
             rightRes[]->left = pRootOfTree;
         }
         return res;
     }

     TreeNode* Convert(TreeNode* pRootOfTree) {
         if(pRootOfTree == nullptr) return nullptr;
         vector<TreeNode* >res =  doConvert(pRootOfTree);
         return res[];
     }

 };

27.输入一个字符串,按字典序打印出该字符串中字符的所有排列。例如输入字符串abc,则打印出由字符a,b,c所能排列出来的所有字符串abc,acb,bac,bca,cab和cba

class Solution {
private:
    void dfs(string str, int k, vector<string>& res) {
        if (k == str.size() - ) {
            res.push_back(str);
            return;
        }

        unordered_set<int> visited;
        sort(str.begin()+k,str.end());
        for (size_t i = k; i < str.size(); i++) {
            if (visited.find(str[i]) == visited.end()) {
                visited.insert(str[i]);
                swap(str[k], str[i]);
                dfs(str, k + , res);
                swap(str[k], str[i]);
            }
        }
    }
public:
    vector<string> Permutation(string str) {
        vector<string> res;
        dfs(str, , res);
        return res;
    }
};

28.数组中有一个数字出现的次数超过数组长度的一半，请找出这个数字。例如输入一个长度为9的数组{1,2,3,2,2,2,5,4,2}。由于数字2在数组中出现了5次，超过数组长度的一半，因此输出2。如果不存在则输出0。

class Solution {
public:
    int MoreThanHalfNum_Solution(vector<int> numbers) {
        int n = numbers.size();
        if(n == ) return ;

        int num = numbers[],count = ;
        for(int i=;i<n;i++){
            if(numbers[i] == num) count++;
            else count--;
            if(count == ){
                num = numbers[i];
                count = ;
            }
        }
        //verify
        count = ;
        for(int i=;i<n;i++){
            if(numbers[i] == num) count++;
        }
        if(count* > n ) return num;
        else return ;  // not exist

    }
};

29.输入n个整数，找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字，则最小的4个数字是1,2,3,4,。solution1在牛客网上超时，solution2能正常通过。

 class Solution1 {
 private:
     int partition(vector<int>& input, int left, int right) {
         if (left >= right)
             return left;
         int pivotVal = input[left];
         while (left < right) {
             while (left < right && input[right] >= pivotVal) {
                 right--;
             }
             input[left] = input[right];
             while (left < right && input[left] <= pivotVal) {
                 left++;
             }
             input[right] = input[left];
         }
         input[left] = pivotVal;
         return left;
     }
 public:
     vector<int> GetLeastNumbers_Solution(vector<int>& input, int k) {
         vector<int> res;
         int n = input.size();

         int index = partition(input, , n - );
         while (index != k - ) {
             if (index > k - ) {
                 index = partition(input, , index - );
             } else {
                 index = partition(input, index + , n - );
             }
         }
         for (int i = ; i < k; i++) {
             res.push_back(input[i]);
         }
         return res;
     }
 };

 class Solution2 {
 public:
     vector<int> GetLeastNumbers_Solution(vector<int>& input, int k) {
         int n = input.size();
         if (n ==  || n < k)
             return vector<int>();
         multiset<int, greater<int>> leastNums(input.begin(), input.begin() + k);

         for (int i = k; i < n; i++) {
             if (input[i] < *leastNums.begin()) {
                 leastNums.erase(leastNums.begin());
                 leastNums.insert(input[i]);
             }
         }
         return vector<int>(leastNums.begin(), leastNums.end());
     }
 };

30HZ偶尔会拿些专业问题来忽悠那些非计算机专业的同学。今天测试组开完会后,他又发话了:在古老的一维模式识别中,常常需要计算连续子向量的最大和,当向量全为正数的时候,问题很好解决。但是,如果向量中包含负数,是否应该包含某个负数,并期望旁边的正数会弥补它呢？例如:{6,-3,-2,7,-15,1,2,2},连续子向量的最大和为8(从第0个开始,到第3个为止)。你会不会被他忽悠住？(子向量的长度至少是1)

class Solution {
public:
    int FindGreatestSumOfSubArray(vector<int>& array) {
        int n = array.size();
        if (n == ) return ;

        int* dp = new int[n];
        dp[] = array[];
        int maxSum = dp[];
        for (int i = ; i < n; i++) {
            dp[i] = max(dp[i - ] + array[i], array[i]);
            if (dp[i] > maxSum) {
                maxSum = dp[i];
            }
        }
        delete [] dp;
        return maxSum;
    }
};