关于网站建设的意义深圳百度seo公司

记录了初步解题思路 以及本地实现代码；并不一定为最优 也希望大家能一起探讨 一起进步

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2/27 1144. 递减元素使数组呈锯齿状

两种情况 第一种奇数位的小 第二种偶数位的小
从头遍历

def movesToMakeZigzag(nums):""":type nums: List[int]:rtype: int"""n = len(nums)ans1,ans2 = 0,0for i in range(n):tmp1,tmp2 = 0,0if i%2==0:if i>0 and nums[i]>=nums[i-1]:tmp1 = nums[i]-nums[i-1]+1if i<n-1 and nums[i]>=nums[i+1]:tmp2 = nums[i]-nums[i+1]+1ans1 += max(tmp1,tmp2)else:if nums[i]>=nums[i-1]:tmp1 = nums[i]-nums[i-1]+1if i<n-1 and nums[i]>=nums[i+1]:tmp2 = nums[i]-nums[i+1]+1ans2 += max(tmp1,tmp2)return min(ans1,ans2)

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2/28 2363. 合并相似的物品

依次遍历 vlist记录所有出现过的价值
集合统计每个价值的重量和

def mergeSimilarItems(items1, items2):""":type items1: List[List[int]]:type items2: List[List[int]]:rtype: List[List[int]]"""m = {}vlist = set()for v,w in items1:m[v] = m.get(v,0)+wvlist.add(v)for v,w in items2:m[v] = m.get(v,0)+wvlist.add(v)ans = []for v in sorted(list(vlist)):ans.append([v,m[v]])return ans

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3/1 2373. 矩阵中的局部最大值

每行每列依次寻找相连三格最大值

def largestLocal(grid):""":type grid: List[List[int]]:rtype: List[List[int]]"""n = len(grid)tmp = [[0]*n for _ in range(n)]for i in range(n):for j in range(1,n-1):tmp[i][j] = max(grid[i][j-1],grid[i][j],grid[i][j+1])ans = [[0]*n for _ in range(n)]for i in range(n):for j in range(1,n-1):ans[j][i] = max(tmp[j][i],tmp[j-1][i],tmp[j+1][i])return [ans[i][1:n-1] for i in range(1,n-1)]

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3/2 面试题 05.02. 二进制数转字符串

小数点后一位一位相加 直至超过32位

def printBin(num):""":type num: float:rtype: str"""curr = 1.0ans = "0."while len(ans)<32:curr = curr/2if num>=curr:num-=currans+="1"else:ans+="0"if num==0:return ansreturn "ERROR"

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3/3 1487. 保证文件名唯一

map存储所有出现过的名字 并记录其当前编号

def getFolderNames(names):""":type names: List[str]:rtype: List[str]"""ans = []m = {}for name in names:if name not in m:ans.append(name)m[name]=1else:ind = m[name]while name+'('+str(ind)+')' in m:ind+=1cur = name+'('+str(ind)+')'ans.append(cur)m[name] = ind+1m[cur] = 1return ans

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3/4 982. 按位与为零的三元组

先统计两数相与能够得到的结果cnt
再依次统计每个数在cnt中有多少能与其相与为0

def countTriplets(nums):""":type nums: List[int]:rtype: int"""cnt = [0]*(1<<16)for i in nums:for j in nums:cnt[i&j] +=1ans = 0for num in nums:target = num ^ 0xffffs = targetwhile True:ans += cnt[s]s = (s-1)&targetif s==target:breakreturn ans

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3/5

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