341. Flatten Nested List Iterator: 用遞歸的方法把值先全部expand华糖，或者在hasnext中expand出一個(gè)integer，判斷出stack的第一個(gè)值時(shí)integer的時(shí)候就返回true
343. Integer Break: dp問(wèn)題，注意前幾個(gè)case，也就是說(shuō)當(dāng)分解后的乘積小于自身的時(shí)候要取自身的值
347.\ Top K Frequent Elements: 用heap自然可以直接做，但是如果在followup里灶体，可以用bucketsortbucketsort的想法就是把同一性質(zhì)的數(shù)放到同一個(gè)bucket里阅签。這里是把具有同樣frequency的數(shù)（先做hash，count frequency）放到bucket里bucket[index] -> list然后由高位到低位依次數(shù)蝎抽。
348. Design Tic-Tac-Toe： 這題還算是簡(jiǎn)單政钟，主要是減少記錄的信息，不需要記錄整個(gè)棋盤(pán)樟结，只要在每下一步养交，記錄一下行，列瓢宦，對(duì)角線(xiàn)碎连，反對(duì)角線(xiàn)的個(gè)數(shù)就好了。
351. Android Unlock Patterns: 一道backtracking的題目驮履，其中的狀態(tài)變化是從某個(gè)按鍵鱼辙，能否移動(dòng)到下一個(gè)按鍵。

class Solution(object):
    def numberOfPatterns(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        skip = [[0 for _ in range(10)] for _ in range(10)]
        skip[1][3] = skip[3][1] = 2
        skip[1][7] = skip[7][1] = 4
        skip[3][9] = skip[9][3] = 6
        skip[7][9] = skip[9][7] = 8
        skip[1][9] = skip[9][1] = skip[2][8] = skip[8][2] = skip[3][7] = skip[7][3] = skip[4][6] = skip[6][4] = 5
        vis = [False]*10
        rst = 0;
        # DFS search each length from m to n
        for i in range(m, n+1):
            rst += self.dfs(vis, skip, 1, i - 1) * 4;    # 1, 3, 7, 9 are symmetric
            rst += self.dfs(vis, skip, 2, i - 1) * 4;    # 2, 4, 6, 8 are symmetric
            rst += self.dfs(vis, skip, 5, i - 1)         # 5
        return rst

    def dfs(self, vis, skip, cur, remain): 
        if remain < 0:
            return 0
        if remain == 0:
            return 1
        vis[cur] = True
        rst = 0
        for i in range(1, 10):
            # If vis[i] is not visited and (two numbers are adjacent or skip number is already visited)
            if not vis[i] and (skip[cur][i] == 0 or vis[skip[cur][i]]):
                rst += self.dfs(vis, skip, i, remain - 1)
        vis[cur] = False
        return rst

353. Design Snake Game: 這種設(shè)計(jì)題不難玫镐，但是要用最精簡(jiǎn)的數(shù)據(jù)結(jié)構(gòu)來(lái)構(gòu)造最合適的倒戏，還是不容易的，可能會(huì)設(shè)計(jì)出不少冗余出來(lái)恐似。
355. Design Twitter：getfeeds 有點(diǎn)類(lèi)似first k element in n sorted linkedlist杜跷，用heap來(lái)做
356. Line Reflection：先按照y值進(jìn)行hash，然后先對(duì)第一個(gè)key找它的reflect line矫夷，然后對(duì)后面所有的值驗(yàn)證一下reflect line
357. Count Numbers with Unique Digits：比較簡(jiǎn)單的一道dp題葛闷，推導(dǎo)公式為dp[i] = dp[i-1] * (10-i+1)
360. Sort Transformed Array：先對(duì)整個(gè)arr進(jìn)行數(shù)學(xué)運(yùn)算，然后從兩頭依次找就可以了双藕。只是要注意a的取值孵运，a是正負(fù)或者是0