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Note: The use of Vector: 1.initialize int x[10]={9,8,7,6,5,4,3,2,1,0}; vector test(x,x+10); 2.end();begin(); vector.begin() points to the first item,while vector.end() doesn't point to the

Description: Given a linked list, remove the nth node from the end of list and return its head. Note: 需要注意的边界情况： 当需要删除的结点是head结点时，head会发生改变。 当n超过结点个数时，表明没有需要删除的结点。 Code: /** * Definiti

Description: Implement an algorithm to delete a node in the middle of a singly linked list, given only access to that node. Note: 1.拿到的是当前的结点，也就是说拿不到当前结点前一个结点，所以我把当前结点用下一个结点的值覆盖掉，再删掉下一个结点。

Description: You have two every large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1. Notice: A tree T2 is a su

跟merge intervals差不多的，没什么好说的。 class Solution { public: /** * Insert newInterval into intervals. * @param intervals: Sorted interval list. * @param newInterval: new interval.

Problem:Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. Solution:采用stack，当字符为上括号时push进stack，否之则从stack pop出字符看是否匹配。

Problem:Given a binary tree, return all root-to-leaf paths. Solution:每个结点记录从root至其的path，当结点无左右孩子时，将path传入进vector。 这里需要注意的是，recordPath函数中，path记录着root至结点的path，不可改变，不能加&（reference），而pathResul

Problem: Given a collection of intervals, merge all overlapping intervals. Solution:先按interval.start按升序进行冒泡排序，再逐一进行merge（如i1和i2比较，若可以merge再i1和后续比较，不可以merge则i2和i3等比较merge）。 Problem:冒泡排序耗时太长。 Mod

直接用了序贯算法写。然后发现！太麻烦了啊啊啊啊啊啊一道easy题写的心好累 （虽然时间复杂度挺好的。 class Solution { public: /** * @param grid a boolean 2D matrix * @return an integer */ int checkEqualList(vector >& equalL

Description: Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target. Note: 1.a valid path的定义是 from root node to any of the leaf nodes.所以当满足target条