度小满

• 机器学习研发

• 选择 30，编程 2

Index

• 火车站台

• 商品交易

火车站台

思路

• 求区间最大重叠数

暴力法（36%）

n = int(input())

tmp = dict()
mx = 0
for _ in range(n):
    x, y = list(map(int, input().split()))
    for i in range(x, y):
        if i in tmp:
            tmp[i] += 1
        else:
            tmp[i] = 1
        mx = max(tmp[i], mx)

print(mx)

扫描线法（AC）

n = int(input())

tmp = []
mx = 0
for _ in range(n):
    x, y = list(map(int, input().split()))
    tmp.append((x, 1))
    tmp.append((y, -1))

tmp.sort()

mx = 0
t = 0
for i in tmp:
    if i[1] == 1:
        t += 1
    else:
        t -= 1

    mx = max(mx, t)

print(mx)

商品交易

思路

• LeetCode原题

• 因为这里要求输出最小交易次数，所以贪心不可行，改用双指针

贪心（9%）

n = int(input())

nums = list(map(int, input().split()))

def foo(nums):
    ans = 0
    cnt = 0
    if len(nums) <= 1:
        return 0
    for x in range(1, len(nums)):
        if nums[x] - nums[x - 1] >= 0:
            ans += nums[x] - nums[x - 1]
            cnt += 2
    return ans, cnt

ans, cnt = foo(nums)
print(ans, cnt)

双指针（AC）

n = int(input())

nums = list(map(int, input().split()))

def foo(nums):
    ans = i = 0
    cnt = 0
    while i < len(nums):
        while i < len(nums) - 1 and nums[i + 1] <= nums[i]:
            i += 1
        minima = nums[i]
        i += 1

        while i < len(nums) - 1 and nums[i + 1] >= nums[i]:
            i += 1

        if i < len(nums):
            ans += nums[i] - minima
            cnt += 2

    return ans, cnt

ans, cnt = foo(nums)
print(ans, cnt)