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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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5/27 2028. 找出缺失的观测数据

算出当前的总和sum(rolls)
算出m+n次的总和mean*(m+n)
后者减去前者就是n次总和s
如果s<n 或者s>6*n 那么不可能存在
先假定n次都是1
再从头开始将1变成6 如果剩余值不能变六就变成剩余值 结束操作

def missingRolls(rolls, mean, n):""":type rolls: List[int]:type mean: int:type n: int:rtype: List[int]"""s = mean*(n+len(rolls))-sum(rolls)if s<n or s>6*n:return []ans =[1]*ns-=nloc = 1while s:if s>=5:ans[loc]+=5loc+=1s-=5else:ans[loc]+=sbreakreturn ans

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5/28 2951. 找出峰值

遍历数组
判断当前位置i的值是否大于左右相邻的值

def findPeaks(mountain):""":type mountain: List[int]:rtype: List[int]"""ans = []for i in range(1,len(mountain)-1):if mountain[i]>mountain[i-1] and mountain[i]>mountain[i+1]:ans.append(i)return ans

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5/29 2981. 找出出现至少三次的最长特殊子字符串 I

记录每个字符最长的三个长度a>=b>=c
要满足三次出现可以有三种情况a-2,min(a-1,b),c

def maximumLength(s):""":type s: str:rtype: int"""from collections import defaultdictl = defaultdict(list)num = 0for i,c in enumerate(s):num+=1if i+1==len(s) or c!=s[i+1]:l[c].append(num)num=0ans = 0for a in l.values():a.sort(reverse=True)a.extend([0,0])ans = max(ans,a[0]-2,min(a[0]-1,a[1]),a[2])return ans if ans else -1

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5/30 2982. 找出出现至少三次的最长特殊子字符串 II

与2981相同
记录每个字符最长的三个长度a>=b>=c
要满足三次出现可以有三种情况a-2,min(a-1,b),c

def maximumLength(s):""":type s: str:rtype: int"""from collections import defaultdictl = defaultdict(list)num = 0for i,c in enumerate(s):num+=1if i+1==len(s) or c!=s[i+1]:l[c].append(num)num=0ans = 0for a in l.values():a.sort(reverse=True)a.extend([0,0])ans = max(ans,a[0]-2,min(a[0]-1,a[1]),a[2])return ans if ans else -1

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5/31 2965. 找出缺失和重复的数字

遍历统计每个数出现次数

def findMissingAndRepeatedValues(grid):""":type grid: List[List[int]]:rtype: List[int]"""n = len(grid)m = [0]*(n*n+1)for row in grid:for v in row:m[v]+=1ans = [0,0]for i in range(1,n*n+1):if m[i]==2:ans[0]=ielif m[i]==0:ans[1]=ireturn ans

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6/1 2928. 给小朋友们分糖果 I

只有三个小朋友 枚举第一个小朋友i个 第二个小朋友j个 第三个小朋友n-i-j个

def distributeCandies(n, limit):""":type n: int:type limit: int:rtype: int"""ans = 0for i in range(limit+1):for j in range(min(n-i,limit)+1):if n-i-j<=limit:ans+=1return ans

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6/2

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