开刷LeetCode，开贴自我督促～有兴趣者可加入

刚做完amazon OA。最近在努力刷面经。。

Compare Version Numbers	理解好题意。不是直接比大小，是以点为间隔，取出每个间隔中的数字进行大小比较。先比较长度大小，相等情况下，再进行每位上数字大小的比较。tricky。	参考曹神的办法做的。很巧妙的算法
Count and Say	没读一个数字跟下一个比较，相同则个数＋＋，不同则继续。在返回的string里只数出个数和数字	改变string要用 str+=**;

最近一直刷题，但是可能就不坚持每天往地里更新了，隔几天更新一次。。currently40/178

Climbing Stairs	same as fibonacci number: 1) recursive ->2) dp ->3)use three variable to save space
Implement strStr()	brute Force.两层循环。	需要找到的needle如果为空，则返回0. 若haystack为空活着小于needle长度返回－1。 确定何时退出。break/continue 区别。
Valid Parentheses	push to stack when ({[ , pop and check when )}]	input something random at first, in case check ] … at first
Set Matrix Zeroes	use row 0 and column 0 to store the result. if(matrix[i][j] == 0) then matrix[i][0] = 0, matrix[0][j] = 0	rows 0 and columns 0 need to be a separate case. Find if they have 0 or not. Assign value 0 later
Find Minimum in Rotated Sorted Array ||	if a[left] > a[right], it is ordered, return a[left]. if a[left] < a[right], find whether left or right part is organized(middle).

Searched in rotated array	binary search. only half is sorted, the other half is not. so need to decide it is in the sorted or unsorted side	the condition of stopping while loop. what idx should be left or right
Unique Paths	have a m*n matrix to store the number of unique paths. Dynamic programming	assign value 1 to 0 array and 0 column.
Unique Paths II	just compute to the given matrix, if it is 1. then dp[j] should be 0	the 0 row should not be 1 any more. also it should be dp[j] = dp[j-1] if the given matrix at this point is 1

最近忙着面Amazon，各种刷面经去了，也都没来的及更新帖子。面完以后就去准备期中考试。空白了两个周。。

现在回归刷题。。目前进度55/181.。。。Leetcode数目这是在不断增加的节奏。跪。

我打算一天刷一道吧。C写得比较熟，打算转java。