3.11
- two-sum (easy)
Q: Given exactly one solution, -> 1 there still can be duplicates 2. exclude that only one of duplicate is the result, which means that both of them or none of them being the result
A: if there is duplicate, check whether these two integers add up to be the result first, after that, we can add it to hashmap. Briefly, just changing the order is enough.
- reverse-integer (easy) (overflow question)
Q: Given a 32-bit signed integer -> really hard to get every digit and hard to deal with the negative part(wrong)
A: feeling hard because you want to get all digits immediately but you can get it one by one using %10 and /10
Q: this is signed integer, and worry about overflow -> just check the final value with Integer.MAX_VALUE (wrong, already overflow)
A: check one step before final step; when dealing with digit, think of 10 into binary (around 2^3);
better way to think about overflow, the key part is: after overflow, the final step cannot be recovered back to previous value
palindrome-number (easy)
int rev = 0;
while(x > rev) {
rev = rev * 10 + x % 10;
x /= 10;
}
return (rev == x) || (rev/10 == x);roman-to-integer (easy)
this is not a normal traverse, it is about combine two elements or not, so we use while(i<len) i++ instead of for loop
3.27
longest-common-prefix (easy)
给一个都是string的array，去找最大的prefix，要注意所需要遍历到的最大index的位置，是不断被更新的，是动态的。squares-of-a-sorted-array (easy)
主要是if else大赏，当需要maintain两个Pointer的时候，就该用while了。toeplitz-matrix (easy)
matrix的行是matrix.length，列是matrix[0].length
3.28
hamming-distance (easy)
奇数和偶数的区别，就是二进制形式最末一位是0还是1。另外右移，就是除以2，形式是a = a >> 1;merge-two-binary-trees (easy)
这里merge的是val，而且要求从root开始merge，所以用preorder。递归的时候注意，return是要return很多次的，这样往左走到头了才能往别的方向跑。
卡了很长时间的地方：if(right...)然后，不是else if(left...)而是if(left...)。 不要乱用else if。single-number (easy)
从都是两个的重复的数中，找出单独的那个数。就是一个小trick，a^a=0。
3.31
- increasing-order-search-tree (easy)
把BST按照inorder重新排列成丿。
要点一：newRoot.right = new TreeNode(root.val) 接续下一个Node的时候，我们要的只是root的value，而不能直接写root，因为那还包括人家自己的left,right指针
要点二:一开始设一个dummy node，作为最终返回的结果。这样子我们就可以像链表一样自我推进，newRoot = newRoot.right;
2.move-zeroes (easy)
inplace地把0移到array右边，同时不影响非0元素的顺序。
好方法：inplace的时候，就利用数组自己的index来存储，因为原数组可能含0，所以把非零数尽可能往数组的前面的Index堆积，再回头补上0就可以。
3.valid-anagram (easy)
这里要尝试考虑unicode characters的进阶：换成hashmap，但是需要自己设定一个0的基准数。(因为数组的优势是自己默认值就是0，这里要考虑同一个字符可能会不止一个)
这里注意遍历的时候，map出来的值是Integer不是Int。
