2019字节跳动算法题

算法题笔记

前期没有练过算法和优化方面的题，所以笔试的时候，做的特别不好，只有第一道题是最简单的。

1.得到的最少硬币数
货币系统包括1、4、16、64元共计4种硬币，以及面值为1024的纸币。现在使用1024的纸币购买一件价值为N(0<N<=1024)的商品，问最少收到多少硬币。
解：
最笨的办法就是(1024-N)/64得到的余数再除以16得到余数再除以4得到余数相加。

class Solution:
    def minnum(self, N):
        num = 0
        res = 1024-N
        for i in [64, 16, 4, 1]:
            num += res/i
            res = res%i
        return num
f = Solution()
N = 200
a = f.minnum(N)
print a
17

3.最少奖品数
有N个人参加比赛，每个人比赛结束后都会得到一个分数，现在将N个人排成一圈领取奖品，要求：
如果某个人的分数比左右的人稿，那么奖品书一定要比左右人的多。
每个人至少得到一个奖品。
问最少应该准备多少奖品？
解：左右遍历，判断奖品数。需要注意的一点是：**N个人围成一个圈。**所以首尾需要判断。

import numpy as np    
class Solution:
    def minnum(self, data):
        n = len(data)
        m = min(data)
        loc = data.index(m)
        res = 0
        #初始化奖品数，保证每个人都有一个奖品
        num1 = [i for i in range(n)]
        num2 = [i for i in range(n)]
        #向右遍历
        for i in range(loc+1, n):
            if data[i] > data[i-1]:
                num1[i] = num1[i-1]+1
        if data[0] > data[n-1]:
            num1[0] = num1[i]
        for i in range(1, loc):
            if data[i] > data[i-1]:
                num1[i] = num1[i-1]+1
        #向左遍历
        for j in range(loc, 0, -1):
            if data[j-1] > data[j]:
                num2[j-1] = num2[j]+1
        if data[n-1] > data[0]:
            num2[n-1] = num2[0] + 1
        for j in range(n-1, loc, -1):
            if data[j-1] > data[j]:
                num2[j-1] = num2[j]
        print num1
        print num2
        for k in range(n):
            res += max(num1[k], num2[k])
        return int(res)

data = [1, 2]
f = Solution()
res = f.minnum(data)

data = [1, 2, 3, 1]
f = Solution()
res = f.minnum(data)           
print res
         
data = [1, 0, 2]
f = Solution()
res = f.minnum(data)           
print res   

data = [1, 2, 2]
f = Solution()
res = f.minnum(data)           
print res   

相似题目Leetcode的题135——分发糖果
描述：
There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:
Each child must have at least one candy.
Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?
解：需要注意的一点是排成一列。所以向左向右判断一遍即可

class Solution(object):
    def candy(self, ratings):
        """
        :type ratings: List[int]
        :rtype: int
        """
        n = len(ratings)
        num1 = [1 for i in range(n)]
        num2 = [1 for i in range(n)]
        res = 0
        for i in range(1,n):
            if ratings[i] > ratings[i-1]:
                num1[i] = num1[i-1]+1
        for j in range(n-1, 0, -1):
            if ratings[j-1] > ratings[j]:
                num2[j-1] = num2[j]+1
        for i in range(n):
            res += max(num1[i], num2[i])
        return res
    
data = [1,2,87,87,87,2,1]
f = Solution()
res = f.candy(data)
print res 
13

4.M根绳子最长长度
有N根绳子，第i根绳子的长度为Li，现在需要M根等长的绳子，可以对N根绳子进行任意裁剪但不能拼接，那么这M根绳子的最长长度是多少。
解：
二分法，逼近求解区间得到最优解

class Solution:
    def maxlength(self, data):
        N = data[0][0]
        M = data[0][1]
        L = data[1] 
        max_length = float(max(L))
        min_length = 0.0
        while(abs(max_length - min_length) >= 1e-3):
            mid = min_length + float(max_length - min_length)/2.0
            num = 0
            for i in range(N):
                num += int(L[i]/mid)
            if num >= M:
                min_length = mid
                res = mid
            else:
                max_length = mid
        return res

data中data[0]表示绳子的个数N和需要绳子个数M
data[1]表示N根绳子的长度；

f = Solution()
a = f.maxlength(data)
data=[[3, 4], [3, 5, 4]]
a = f.maxlength(data)
a
Out[16]: 2.5
data = [[3, 6], [5, 2, 3]]
a = f.maxlength(data)
a
Out[19]: 1.4996337890625
data = [[4, 5], [1, 2, 3, 4]]
a = f.maxlength(data)
a
Out[22]: 1.5