markpen的专栏

leetcode 324 wiggle sort II

Leetcode 215 Kth Largest Element in an Array

Leetcode 260 SIngle Number III

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 207 Course Schedule

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：找到一棵树中，各结点值和最大的路径。思路：DFS， 返回必须经过该结点的路径的最大值和可能经过该结点的路径的最大值。/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * Tr

题意：给出地形的高度，求出能汇集的最多的雨水。思路：这题和什么时候买股票那题有点像。数组lr记录到目前为止，左边的最高高度；数组rl记录到目前为止，右边的最高高度。他们的最小值就是能汇集的水量。class Solution {public: int trap(vector& height) { if(height.size() <= 1) return 0;

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：给出一数组，求i到j之间的和。题意：记录开始位置到各位置的和，查询时，减去开头部分的和。class NumArray {public: NumArray(vector &nums) { totalSum.push_back(0); int sum = 0; for(int i = 0; i < nums.size(); ++

题意：删除链表中倒数第i个元素。思路：用堆栈完成计数工作。/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class So

题意：求最长回文子串。思路：记录到目前为止，以该字母结尾的所有字串的起始位置，之后的字符遍历这些长度，记录更新后的起始位置。class Solution {public: string longestPalindrome(string s) { vector > pal; vector temp(1, 0); pal.push_ba

题意：求一个平面中，一个点到其他两个点的距离相等的点，这样的点一共有多少组。思路：求出所有两两点之间的距离，用hash统计距离相同的点。class Solution {public: int numberOfBoomerangs(vector>& points) { vector > distance; for(int i = 0