刚看到本题时感觉难度不是很大，却出现在困难组里面。后面万万没想到提交测试的时候输入了将近20000条直线进行判断，如果直接上来就做的话，那肯定会：
接下来看题目：
题目
python代码
刚开始我的代码是直接做的，结果超时。
def minRecSize(lines):
"""
:type lines: List[List[int]]
:rtype: float
"""
x_list=[]
y_list=[]
for i in range(len(lines)):
for j in range(i + 1, len(lines)):
k1, b1 = lines[i]
k2, b2 = lines[j]
if k1 != k2:
x = (b2 - b1) / (k1 - k2)
y = ((b2 - b1) / (k1 - k2)) * k1 + b1
x_list.append(x)
y_list.append(y)
if len(x_list)<2:
return 0
x_list.sort()
y_list.sort()
wideth=x_list[-1]-x_list[0]
longth=y_list[-1]-y_list[0]
area=wideth*longth
# print(x_list)
# print(y_list)
# print(area)
return area原因就是我的做法将所有斜率不同的线两两求交点，这样的排列组合在两万条线的情况下……
所以要先处理输入的数据，做法：将所有斜率相同的直线，保留截距最大和最小的那两条，处理所有的直线，将处理后的结果存到新的list中作为上面代码的输入。
def deal(lines):
k_dict={
}
new_lines=[]
for k,b in lines:
print(k,b)
if k in k_dict:
k_dict[k][0]=min(k_dict[k][0],b)
k_dict[k][1]=max(k_dict[k][1],b)
else:
k_dict[k]=[b,b]
print(k_dict)
for key,val in k_dict.items():
new_lines1=[]
new_lines2=[]
k=key
b1=val[0]
b2=val[1]
new_lines1.append(k)
new_lines1.append(b1)
new_lines2.append(k)
new_lines2.append(b2)
if new_lines1 not in new_lines:
new_lines.append(new_lines1)
if new_lines2 not in new_lines:
new_lines.append(new_lines2)
# print(len(new_lines))
return new_lines
def minRecSize(lines):
new_lines=deal(lines)
print(new_lines)
x_list=[]
y_list=[]
for i in range(len(new_lines)):
for j in range(i + 1, len(new_lines)):
k1, b1 = new_lines[i]
k2, b2 = new_lines[j]
if k1 != k2:
x = (b2 - b1) / (k1 - k2)
y = ((b2 - b1) / (k1 - k2)) * k1 + b1
x_list.append(x)
y_list.append(y)
if len(x_list)<2:
return 0
x_list.sort()
y_list.sort()
wideth=x_list[-1]-x_list[0]
longth=y_list[-1]-y_list[0]
area=wideth*longth
return area继续超出时间限制
接着看到别人更简洁的代码，思路一致，但是省去了一些比较、排序所费的时间
def minRecSize(self, lines: List[List[int]]) -> float:
if len(lines) <= 2:
return 0
dic = {
}
for k, b in lines:
if k in dic:
dic[k][0] = min(dic[k][0], b)
dic[k][1] = max(dic[k][1], b)
else:
dic[k] = [b, b]
if len(dic) == 1:
return 0
ks = sorted(dic.keys())
n = len(ks)
min_x = min_y = float('inf')
max_x = max_y = float('-inf')
for i in range(1, n):
k1 = ks[i - 1]
min_b1, max_b1 = dic[k1]
k2 = ks[i]
min_b2, max_b2 = dic[k2]
diff_k = k1 - k2
min_x = min(min_x, (max_b2 - min_b1) / diff_k)
max_x = max(max_x, (min_b2 - max_b1) / diff_k)
min_y = min(min_y, (k1 * max_b2 - k2 * min_b1) / diff_k)
max_y = max(max_y, (k1 * min_b2 - k2 * max_b1) / diff_k)
return (max_x - min_x) * (max_y - min_y)