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  1. 200. Number of Islands

    class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        newgrid = [list(x) for x in grid]
        ans = 0
        if not len(newgrid):
            return ans
        m = len(newgrid)
        n = len(newgrid[0])
        for i in range(m):
            for j in range(n):
                if newgrid[i][j ...
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  2. 199. Binary Tree Right Side View

    # Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def rightSideView(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        ans = []
        queue = [root]
        while queue:
            for i in range(len(queue ...
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  3. 198. House Robber

    class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        #dp[i] = max(dp[i-1], dp[i-2] + nums[i])
        if not nums:
            return 0
        if len(nums) <= 2:
            return max(nums)
        dp = [0] * len(nums)
        dp[0] = nums[0]
        dp[1] = max(nums[0], nums[1])
        i ...
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  5. 190. Reverse Bits

    # -*- coding:utf8 -*-
class Solution:
    # @param n, an integer
    # @return an integer
    def reverseBits(self, n):
        new = 0
        for i in range(32):
            new = (new << 1) | ((n >> i) % 2)
            # n右移并%2得到二进制表达下的每位值。然后，new左移得到xxx0，'|' n的每位值，即xxx0可能变成xxx1。
        return new

if __name__ == "__main__":
    print Solution().reverseBits(43261596)
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  6. 189. Rotate Array

    class Solution(object):
    def rotate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        # san bu fanzhuan fa
        k %= len(nums)
        if k == 0:
            return
        start1= 0
        end1 = len(nums)-k-1
        for i in range((end1-start1)/2 ...
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  7. 187. Repeated DNA Sequences

    class Solution(object):
    def findRepeatedDnaSequences(self, s):
        """
        :type s: str
        :rtype: List[str]
        """
        n = len(s)
        d = {}
        res = []
        for i in range(n):
            substr = s[i:i + 10]
            d[substr] = d.get(substr, 0) + 1
        for key, val in d.items():
            if val > 1:
                res.append(key)
        return res
if ...
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  8. 186. Reverse Words in a String II

    # Time: O(n)
# Space:O(1)
#
# Given an input string, reverse the string word by word.
# A word is defined as a sequence of non-space characters.
#
# The input string does not contain leading or trailing spaces
# and the words are always separated by a single space.
#
# For example,
# Given s ...
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  9. 179. Largest Number

    # -*- coding:utf8 -*-
class Solution:
    # @param {integer[]} nums
    # @return {string}
    def largestNumber(self, nums):
        res = ''
        while nums:
            item = nums.pop(self.help(nums))
            res += str(item)
        if (res[0]=='0'):
            return '0'
        return res

    def help(self, nums):  # 如果a＋b串大于b＋a串，那么a比较大，反之b比较大。
        strnums = [str(x) for x in nums]
        index ...
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