前两题有点简单,是签到题,就不耽误大家时间了
第三题主要用hashmap存每个分数的次数, 1 / 2 1/2 1/2 取出来做特殊判断,对去重后的分数进行排序,然后用双指针遍历排序后的分数,找到和为1的两个数。
AC代码:
def solution(X, Y):
n = len(X)
div = []
counter = {
}
for i in range(n):
x, y = X[i], Y[i]
f = x / y
# 考虑只有1/2的特殊情况
counter[f] = counter.get(f, 0) + 1
if counter[f] == 1 and f != 0.5:
div.append((f, i))
div.sort()
l, r = 0, len(div) - 1
k = counter.get(.5, 0)
ans = k * (k - 1) // 2
while r > l:
# 判断相加为1 todo: 化简(虽然python不会因为大数而报错)
a, b = X[div[l][1]], Y[div[l][1]]
c, d = X[div[r][1]], Y[div[r][1]]
if a * d + b * c == b * d:
ans += counter[div[l][0]] * counter[div[r][0]]
l += 1
r -= 1
elif div[l][0] + div[r][0] > 1:
r -= 1
else:
l += 1
return ans % int(1e9 + 7)
def gcd(a,b):
if a%b == 0:
return b
else :
return gcd(b,a%b)
print(solution([1, 2, 3, 1, 2, 12, 8, 4], [5, 10, 15, 2, 4, 15, 10, 5]))
print(solution([1, 1, 1], [2, 2, 2]))
print(solution([1, 1, 2], [3, 2, 3]))
print(solution([2, 1, 1], [4, 3, 2]))