markpen的专栏

pratiseEasy Game#include #include #include #include using namespace std;int main(){ int n,cas=0; cin>>n; string s; while(n--) { cas++; cin>>s; if

leetcode 324 wiggle sort II

Leetcode 215 Kth Largest Element in an Array

Leetcode 338 Counting Bits

Leetcode 289 Game of Life

Leetcode 73 Set Matrix Zeros

Leetcode162 Find Peak Element

Leetcode 119 Pascal's Triangle II

Leetcode 200 Number of Islands

Leetcode 274 H-Index

Leetcode 275 H-Index II

leetcode 383 Ransom Note

Leetcode 104 Maximum Depth of Binary Tree

Leetcode 226 Invert Binary Tree

Leetcode 83 Remove Duplicates from Sorted List

Leetcode 53 Maximum Subarray

Leetcode 89 Gray Code

题意：判断一棵树是否为平衡树。平衡树是指，各子树的高度之差不超过1。思路：计算各子树高度，并判断。class Solution {public: bool isBalanced(TreeNode* root) { if(!root) return true; bool bleft = false; bool bright = fal

listclass Solution {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode *temp1 = l1; ListNode *temp2 = l2; ListNode *head = new ListNode(0); head->nex

题意：输入数字，找所以可能的字母组合。思路：DFS，到达叶结点输出。class Solution {public: vector b; vector re; vector letterCombinations(string digits) { string s = "abc"; b.push_back(s); s = "def";

题意：括号匹配。思路：堆栈。class Solution {public: bool isValid(string s) { stack mys; mys.push(s[0]); for(unsigned i = 1; i < s.length(); ++ i) { if(!mys.empty()) {

题意：合并两个排列好的队列。思路：列表的基本操作。class Solution {public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode *nextl1 = l1; ListNode *nextl2 = l2; ListNode

题意：写出所有合法的括号组合。思路：DFS。class Solution {public: vector re; vector generateParenthesis(int n) { string s = ""; dfs(s, n, n); return re; } int dfs(str

题意：检查数独是否合法，行列及九宫格的数字 不能重复。思路：暴力检查。class Solution {public: bool isValidSudoku(vector>& board) { bool isRepeat[10]; for(int i = 0; i < board.size(); ++ i) {

题意：整数加一。思路：模拟题。class Solution {public: vector plusOne(vector& digits) { vector re; int c = 0; for(int i = digits.size(); i > 0; -- i) { int d = digits[i -

题意：判断两棵树是否相等。思路：DFS。class Solution {public: bool isSameTree(TreeNode* p, TreeNode* q) { if(p == NULL && q == NULL) return true; if(p == NULL && q != NULL) return false;

题意：只能向下或者向右，求路径最短。思路：简单DP。class Solution {public: int minPathSum(vector>& grid) { int m = grid.size(); int n = grid[0].size(); vector > dis = grid; fo

题意：判断是否存在从根结点到叶结点，路上结点之和为所给的值的路径。思路：DFS。class Solution {public: bool hasPathSum(TreeNode* root, int sum) { if(root == NULL) return false; sum -= root->val; bo

题意：找树最浅的深度。思路：DFS。class Solution {public: int minDepth(TreeNode* root) { if(root == NULL) return 0; if(root->left == NULL && root->right == NULL) return 1; int deeplef

题意：编写一个基本的计算器，实现加减法操作。思路：用堆栈完成优先级操作。class Solution {public: int calculate(string s) { stack cal; for(int i = 0; i < s.length(); ++ i) { if(s[i] == ' ') continue;

题意：用链表实现插入排序。思路：简单模拟，注意处理链表的开头与结尾。/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */cla

题意：在矩阵中找最大递增路径。思路：DFS。向可能的四个方向搜索，将返回的最大值 + 1作为该结点的值。class Solution {public: vector > m; int xx,yy; vector > mmatrix; int longestIncreasingPath(vector>& matrix) { if(matrix

题意：按字符出现频率降序输出字符。思路：用hash实现O(1)插入和查找。class Solution {public: string frequencySort(string s) { map m; for(int i = 0; i < s.length(); ++ i) { m[s[i]] ++; }

题意：将一个数字转化成十六进制的形式。思路：简单模拟，注意负数符号位的处理。class Solution {public: string toHex(int num) { string re; if(num == 0) re += '0'; while(num) { int temp = num & 15;

题意：给出股票价格，可以多次买入、卖出，求利润的最大值。思路：只要股票价格上涨，就买入卖出。class Solution {public: int maxProfit(vector& prices) { if(prices.size() == 0) return 0; int maxbuy = 0; for(int

题意：找出各结点之和为给定值的所有路径。思路：DFS。/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL)

题意：不使用除法操作完成整除运算。思路：采用和快速幂相类似的做法。注意边界处理。class Solution {public: int divide(int dividend, int divisor) { long df = 0; long tempd = dividend; long tempdi = divisor;

题意：删除列表中重复的所有元素。思路：用hash表记录重复的元素，在链表中删除。/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * };

题意：统计战舰的个数，其实就是统计相连的X的个数。思路：逐个检查，用DFS改相连的X。class Solution {public: vector > tempboard; int sizex, sizey; int countBattleships(vector>& board) { tempboard = board; size

题意：求多少各连续自然数之和小于n。思路：求和公式。class Solution {public: int arrangeCoins(int n) { double tempn, tempx; tempn = n; tempx = sqrt(2 * tempn + 0.25) - 0.5; return floor(