摘要：
This is pretty simple.Just use three element to pretain the result; 1 class Solution { 2 public: 3 int climbStairs(int n) { 4 vector dp(3,... 阅读全文
摘要：
Two notes:1. result at least n candys. Use array scan from beginning and from end.2. dp[i] = max(dp[i], count++). count++ already record the increasin... 阅读全文
摘要：
Made a stupid bug....... When reverse the vector, stop it at mid. Otherwise, it will swap back......Mark!!!!!!!! 1 /** 2 * Definition for binary tree... 阅读全文
摘要：
Recursive method can be unstand easily:1. Assume we get the sub node with this transition. So we need to make the current node.2. As the symmetic, the... 阅读全文
摘要：
Same with inorder.1. recursive: 1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeN... 阅读全文
摘要：
Still 3 methods:Same with inorder, but post order is a little different. We need to treat it as reverse result of preorder.So we go to right branch fi... 阅读全文
摘要：
Use lMax get the maximum value of a sub SINGLE branch (left branch or right branch or just current node).Use gMax get the maximum value of the whole s... 阅读全文
摘要：
This question is almost same as the previous one.I just use a stack (another vector) of vector to store the level result. Then reverse it. 1 /** 2 * ... 阅读全文
摘要：
Basic question. Use a queue to store Node, use two numbers to record current level node numbers and next level node numbers. 1 /** 2 * Definition for... 阅读全文
摘要：
There are three methods to do it:1. recursive(use memory stack): (Time O(n), Space O(logn) 1 /** 2 * Definition for binary tree 3 * struct TreeNode ... 阅读全文
摘要：
It is similar to use stack for BST inorder traversal.So do the same thing :1. Get stack to store right branchs(include current node).2. When you pop t... 阅读全文
摘要：
This is combination of II and III.1. when k >= length, it is totally II.2. when k localMax[j] = max(localMax[j] + diff, globalMax[j-1] + max(0, diff... 阅读全文
摘要：
III is a kind of I. But it require 2 maximum value. So scan from begin and scan from end to record the maximum value for currrent index.Then scan the ... 阅读全文
摘要：
This is simple, use greedy algorithm will solve the problem. 1 class Solution { 2 public: 3 int maxProfit(vector &prices) { 4 int len = pr... 阅读全文
摘要：
1 class Solution { 2 public: 3 int maxProfit(vector &prices) { 4 if (prices.size() < 2) return 0; 5 int len = prices.size(), resu... 阅读全文
摘要：
Need a helper function to return the binary tree maximum height. If the return value is -1, it means there is a imbalanced subbranch . 1 /** 2 * Defi... 阅读全文
摘要：
Use a hash table to record it. The tricky part is that when the hash value is > 0, it is the index + 1. Otherwise, that string already exists in resul... 阅读全文
摘要：
Same with add binary. You can also skip delete the result pointer. But just return result->next. Depends on the interviewer. 1 /** 2 * Definition for... 阅读全文
摘要：
This question is pretty straight forward. 1 class Solution { 2 public: 3 string addBinary(string a, string b) { 4 int runnerA = a.size()-1... 阅读全文
摘要：
To be honest, I dont like this problem. This is just an extra layer loop for 3sum. But two mistakes I had made…..1. when skip j, use condition j > i+1... 阅读全文