碰见这题的时候是一年前了。忙碌的一年下来再次回头撸一下，发现emmmm。。。竟然可以用dp做出来(bushi)，有点欣慰额~~~
# -*- coding: utf-8 -*-
import copy
class Solution(object):
def makesquare(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
if sum(nums) % 4: return False
target = sum(nums) / 4
if all([True if num <= target else False for num in nums]) == False: # 检查是否有火柴大于边长
return False
# 不能用dp ...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 ...# 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 ...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 ...class Solution(object):
def hammingWeight(self, n):
"""
:type n: int
:rtype: int
"""
count = 0
for i in range(32):
if (n>>i)%2 == 1:
count += 1
return count
# -*- 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)
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 ...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 ...# 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 ...Page 1 / 21 »