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
可以将25ml看作1份 四种情况分别为
A 4 B 0
A 3 B 1
A 2 B 2
A 1 B 3
开始时 A,B拥有的份数一样
当n过大时 A被取完的概率接近于1 误差小于10^-5
假设dp[i][j] 为A剩余i份 B剩余j份时 需要求的概率
dp[i][j] = (dp[i-4][j]+dp[i-3][j-1]+dp[i-2][j-2]+dp[i-1][j-3])/4
当j=0,i>0时 dp[i][j] = 0
当i=0,j>0时 dp[i][j] = 1
当i=0,j=0时 dp[i][j] = 0.5
def soupServings(n):""":type n: int:rtype: float"""n = (n+24)//25if n>=179:return 1dp = [[0]*(n+1) for _ in range(n+1)]dp[0] = [0.5]+[1.0]*nfor i in range(1,n+1):for j in range(1,n+1):dp[i][j] = (dp[max(0, i - 4)][j] + dp[max(0, i - 3)][max(0, j - 1)] +dp[max(0, i - 2)][max(0, j - 2)] + dp[max(0, i - 1)][max(0, j - 3)])/4return dp[n][n]
f(x)表示小于等于x内的神奇数字数目 c为a,b最小公倍数lcm
f(x)=x//a+x//b-x//c
f(x)随x增加单调不减 所以可以二分查找第n个神奇数字
def nthMagicalNumber(n, a, b):""":type n: int:type a: int:type b: int:rtype: int"""def lcm(a,b):x,y = a,bif a0:x,y = y,x%yreturn a*b/xMOD = 10**9+7l,r = min(a,b),min(a,b)*nc = lcm(a,b)while l<=r:mid = (l+r)>>1s = mid//a+mid//b-mid//cif s>=n:r = mid-1else:l = mid+1return (r+1)%MOD
用map存储每个盒子内的小球数
def countBalls(lowLimit, highLimit):""":type lowLimit: int:type highLimit: int:rtype: int"""def func(x):tag = 0while x>0:tag += x%10x = x//10return tagdic = {}ans = 0for i in range(lowLimit,highLimit+1):tag = func(i)dic[tag] = dic.get(tag,0)+1if dic[tag]>ans:ans = dic[tag]return ans
子数组最大值不能大于right 且范围内至少有一个处于[left,right]间
l表示上次出现大于right的位置
r表示上次出现left,right之间的位置
对于当前位置i 可以有r-l个选择
遍历每一个位置可以得到答案
def numSubarrayBoundedMax(nums, left, right):""":type nums: List[int]:type left: int:type right: int:rtype: int"""ans = 0l,r = -1,-1for i,num in enumerate(nums):if left<=num<=right:r = ielif num>right:l = ir = -1if r!=-1:ans += r-lreturn ans
func 将字符串中连续相同的字母压缩成一个字母+出现的次数
比较原始s和需要比较的字符串w 压缩后的状态
需要满足 长度相等 每个位置上的字母需要一样
出现的次数 s和w相同 或者s>w 并且s的次数大于等于3
def expressiveWords(s, words):""":type s: str:type words: List[str]:rtype: int"""def func(s):l = []cur = ""num = 0for c in s:if cur=="":cur = cnum=1elif cur==c:num+=1else:l.append((cur,num))cur = cnum = 1l.append((cur,num))return lsl = func(s)ans = 0for w in words:wl = func(w)if len(wl)!=len(sl):continuetag = Truefor i in range(len(sl)):if sl[i][0]!=wl[i][0]:tag = Falsebreakif wl[i][1]==sl[i][1] or (sl[i][1]>wl[i][1] and sl[i][1]>=3):continueelse:tag = Falsebreakif tag:ans+=1return ans