字节跳动2019春招研发部分编程题汇总

1。万万没想到之聪明的编辑
添加链接描述

import sys
n = int(sys.stdin.readline()[:-1])
input = []
for i in range (n):
    input.append(sys.stdin.readline()[:-1])
#print(input)
result = []
for inp in input:
    length = len(inp)
    res = inp[:2]         
    if length >= 3:
        for i in range(2, length):
            #print('inp ', inp)
            #print('res ', res)
            if inp[i] == res[-1] and res[-1] == res[-2]:
                continue
            elif inp[i] == res[-1] and res[-2] == res[-3]:
                continue
            res += inp[i]
    result.append(res)
for res in result:
    print(res)

2.万万没想到之抓捕孔连顺
添加链接描述

import sys
n,d = map(int, input().split())
nums = list(map(int, input().split()))
#nums = (sys.stdin.readline()[:-1].split())
#nums = [int(f) for f in nums]
left, right, res = 0, 2, 0
while left < n-2:
    while right < n and nums[right] - nums[left] <= d:
        right += 1
    num = right - 1 - left
    res += num * (num - 1) // 2  #计算left到right之间的点k的组合数C(k, 2)
    left += 1
print(res % 99997867)

3.雀魂启动！
添加链接描述

def isHu(nums):
    if not nums:
        return True
    n = len(nums)
    count = nums.count(nums[0])
    # 没出现过雀头，且第一个数字出现的次数 >= 2,去掉雀头剩下的能不能和牌
    if n%3 and count >= 2 and isHu(nums[2:]):
        return True
    # 如果第一个数字出现次数 >= 3，去掉这个刻子后看剩下的能和牌
    if count >=3 and isHu(nums[3:]):
        return True
     # 如果存在顺子，移除顺子后剩下的能和牌
    if nums[0]+1 in nums and nums[0]+2 in nums:
        tmp = list(nums)  #保存现场
        tmp.remove(nums[0])
        tmp.remove(nums[0]+1)
        tmp.remove(nums[0]+2)
        if isHu(tmp):
            return True
    return False

def main(nums):
    d = {}  #统计出现次数（词频）
    for i in nums:
        d[i] = d.get(i, 0) + 1
    card_list = set(range(1, 10)) - {k for k, v in d.items() if v == 4}  #只取次数为4的，用来做差集
    res = []
    for i in card_list:
        if isHu(sorted(nums + [i])):
            res.append(i)
    print(' '.join(str(r) for r in sorted(res)) if res else '0')
        

nums = [int(x) for x in input().split()]
main(nums)

4.特征提取
添加链接描述

case = int(input())
res = 0
for _ in range(case):  #遍历每个用例
    frames = int(input())
    last_map = {}    #上一帧频率
    for _ in range(frames):  #遍历每帧
        line = list(map(int, input().split(' ')))  #每行转成数组
        n = line.pop(0)  #特征数
        current_map = {}  #记录特征频率
        for i in range(n):  #取出每帧所有特征
            feature = (line[2*i], line[2*i+1])
            #print(feature)
            current_map[feature] = last_map.get(feature, 0) + 1  #统计特征频率
            res = max(res, current_map[feature])    #记录最大连续帧
        #print(last_map, current_map)
        last_map = current_map    #更新

print(res if res >= 2 else 1)

5.毕业旅行问题
添加链接描述

dfs回溯，没有完全通过

import sys
n = int(input())
m = []
for i in range(n):
    m.append(list(map(int, input().split(' '))))
#print(m)
vis = [False] * n
vis[0] = True
res = sys.maxsize

def abc():
    print(m)
def dfs(vis, local, vn, price):
    global res
    #print(res)
    if price > res:
        return
    if vn == n:
        val = price + m[local][0]
        res = min(res, val)
        return
    for i in range(1, n):
        if vis[i]:
            continue
        vis[i] = True
        dfs(vis, i, vn+1, price + m[local][i])
        vis[i] = False
#abc()
dfs(vis, 0, 1, 0)
print(res)

dp:
添加链接描述

6.找零
添加链接描述

num = 1024 - int(input())
coin_64 = num // 64
num = num % 64
coin_16 = num // 16
num = num % 16
coin_4 = num // 4
num = num % 4
coin_1 = num
print(coin_64 + coin_16 + coin_4 + coin_1)

7.机器人跳跃问题
添加链接描述
一维dp：

input()
arr = list(map(int, input().split(' ')))
# 假设跳跃前能力为E,要跳的高度为H，那么跳跃后的能量就是2E-H，
# 那么跳跃后的能量加上高度就是跳跃前的两倍，然后从后往前逆推。
E = 0    # 跳到最后一步的能力值设为0
#从后往前
#递推公式为E(k) = [E(k+1)+H(k+1)] / 2
#由于E都是整数，因此考虑向上取整，故实际运算时用[E(k+1)+H(k+1)+1] // 2
for H in arr[::-1]:
    E = (E + H + 1) >> 1  #用右移运算代替对2整数除也可
print(E)