Russian Doll Envelopes Largest Divisible Subset Two Sum - Input array is sorted
Russian Doll Envelopes
俄罗斯玩偶嵌套问题，这个是典型的dp问题···强行遍历会提示超时，然而整了好久也没整明白怎么整，网上搜了下 把问题归结为求最长递增子序列问题··然而本人愚钝还是想不明白为啥可以这样做··虽然出来的结果是对的·····
先把数据排序， 用python内建的排序函数进行排序，但是因为当x相等时，y要按从大到小拍，所以要传一个cmp进去，python3.x不支持cmp了 所以 用了一个转换，转换成key，如果直接key设置为x默认y会按从小到大拍
这样算的结果是对的·但是那个迭代的dp不是一个有效的序列···但是长度是对的···
class Solution:
# @param {int[][]} envelopes a number of envelopes with widths and heights
# @return {int} the maximum number of envelopes
def maxEnvelopes(self, envelopes):
# Write your code here
import functools
nums = sorted(envelopes,key= functools.cmp_to_key(lambda x,y:x[0]-y[0] if x[0] != y[0] else y[1] - x[1]))
size = len(nums)
dp = []
for x in range(size):
low, high = 0, len(dp) - 1
while low <= high:
mid = (low + high)//2
if dp[mid][1] < nums[x][1]:
low = mid + 1
else:
high = mid - 1
if low < len(dp):
dp[low] = nums[x]
else:
dp.append(nums[x])
return len(dp)
Largest Divisible Subset
标签写的是动态规划 ·感觉没啥规划还是直接强行遍历的··
class Solution:
# @param {int[]} nums a set of distinct positive integers
# @return {int[]} the largest subset
def largestDivisibleSubset(self, nums):
# Write your code here
n=len(nums)
nums= sorted(nums,reverse=True)
res=[]
res.append([nums[0]])
for i in range(1,n):
cur=nums[i]
for r in res:
if r[-1] % cur== 0:
r.append(cur)
if i==1:res.append([nums[0]])
res.append([nums[i]])
res=sorted(res,key=lambda x:len(x),reverse=True)
return res[0]
Two Sum - Input array is sorted
有序数组，找出一个组合之和是给定目标值，题目写的有序基本就是说用二分查找吧···，而且还要求index2>index1相当简单了··，遍历一遍就可以了··
class Solution:
"""
@param nums {int[]} n array of Integer
@param target {int} = nums[index1] + nums[index2]
@return {int[]} [index1 + 1, index2 + 1] (index1 < index2)
"""
def twoSum(self, nums, target):
# Write your code here
index1=0
index2=-1
for i in range(len(nums)-1):
index1 = i
index2 = -1
start = i+1
end = len(nums)-1
st = target - nums[index1]
while start <= end:
mid = (start + end) // 2
if nums[mid] < st:
start = mid + 1
elif nums[mid] > st:
end = mid - 1
else:
index2 = mid
return [index1 + 1, index2 + 1]