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剑指offer(26-30)编程题

2024-09-11 12:11:46   214人阅读

  • 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指向

class Solution {
public:
    //0:left 1: right
    TreeNode* doConvert(TreeNode* pRootOfTree, int leftOrRight) {
        if (pRootOfTree == nullptr) return nullptr;
        TreeNode* pleft = doConvert(pRootOfTree->left, 0);
        TreeNode* pright = doConvert(pRootOfTree->right, 1);
        if (pleft) pleft->right = pRootOfTree;
        if (pright) pright->left = pRootOfTree;
        pRootOfTree->left = pleft;
        pRootOfTree->right = pright;
        //left,返回最右边的节点
        if (leftOrRight == 0) {
            while (pRootOfTree->right) {
                pRootOfTree = pRootOfTree->right;
            }
        }
        //right,返回最左边的节点
        if (leftOrRight == 1) {
            while (pRootOfTree->left) {
                pRootOfTree = pRootOfTree->left;
            }
        }
        return pRootOfTree;
    }

public:
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (pRootOfTree == nullptr)
            return nullptr;
        TreeNode* pleft = doConvert(pRootOfTree->left, 0);
        TreeNode* pright = doConvert(pRootOfTree->right, 1);
        if (pleft) pleft->right = pRootOfTree;
        if (pright) pright->left = pRootOfTree;
        pRootOfTree->left = pleft;
        pRootOfTree->right = pright;

        //right,返回最左边的节点
        while (pRootOfTree->left) {
            pRootOfTree = pRootOfTree->left;
        }
        return pRootOfTree;
    }
};

 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() - 1) {
            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 + 1, res);
                swap(str[k], str[i]);
            }
        }
    }
public:
    vector<string> Permutation(string str) {
        vector<string> res;
        dfs(str, 0, 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 == 0) return 0;

        int num = numbers[0],count = 1;
        for(int i=1;i<n;i++){
            if(numbers[i] == num) count++;
            else count--;
            if(count == 0){
                num = numbers[i];
                count = 1;
            }
        }
        //verify
        count = 0;
        for(int i=0;i<n;i++){
            if(numbers[i] == num) count++;
        }
        if(count*2 > n ) return num;
        else return 0;  // 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。

 1 class Solution1 {
 2 private:
 3     int partition(vector<int>& input, int left, int right) {
 4         if (left >= right)
 5             return left;
 6         int pivotVal = input[left];
 7         while (left < right) {
 8             while (left < right && input[right] >= pivotVal) {
 9                 right--;
10             }
11             input[left] = input[right];
12             while (left < right && input[left] <= pivotVal) {
13                 left++;
14             }
15             input[right] = input[left];
16         }
17         input[left] = pivotVal;
18         return left;
19     }
20 public:
21     vector<int> GetLeastNumbers_Solution(vector<int>& input, int k) {
22         vector<int> res;
23         int n = input.size();
24 
25         int index = partition(input, 0, n - 1);
26         while (index != k - 1) {
27             if (index > k - 1) {
28                 index = partition(input, 0, index - 1);
29             } else {
30                 index = partition(input, index + 1, n - 1);
31             }
32         }
33         for (int i = 0; i < k; i++) {
34             res.push_back(input[i]);
35         }
36         return res;
37     }
38 };
39 
40 class Solution2 {
41 public:
42     vector<int> GetLeastNumbers_Solution(vector<int>& input, int k) {
43         int n = input.size();
44         if (n == 0 || n < k)
45             return vector<int>();
46         multiset<int, greater<int>> leastNums(input.begin(), input.begin() + k);
47 
48         for (int i = k; i < n; i++) {
49             if (input[i] < *leastNums.begin()) {
50                 leastNums.erase(leastNums.begin());
51                 leastNums.insert(input[i]);
52             }
53         }
54         return vector<int>(leastNums.begin(), leastNums.end());
55     }
56 };

 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 == 0) return 0;

        int* dp = new int[n];
        dp[0] = array[0];
        int maxSum = dp[0];
        for (int i = 1; i < n; i++) {
            dp[i] = max(dp[i - 1] + array[i], array[i]);
            if (dp[i] > maxSum) {
                maxSum = dp[i];
            }
        }
        delete [] dp;
        return maxSum;
    }
};

 

剑指offer(26-30)编程题


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