代码随想录算法训练营 Day 26
Published:
Q1. LeetCode 452
按照气球起始位置正序排列
class Solution:
def findMinArrowShots(self, points: List[List[int]]) -> int:
if len(points) == 0:
return 0
points.sort(key=lambda x:x[0])
res = 1
curr_right = points[0][1]
for i in range(1, len(points)):
if points[i][0] > curr_right:
res += 1
curr_right = points[i][1]
else:
curr_right = min(curr_right, points[i][1])
return res
按照气球终止位置逆序排列
class Solution:
def findMinArrowShots(self, points: List[List[int]]) -> int:
if len(points) == 0:
return 0
points.sort(key=lambda x:-x[1])
res = 1
curr_left = points[0][0]
for i in range(1, len(points)):
if points[i][1] < curr_left:
res += 1
curr_left = points[i][0]
else:
curr_left = max(curr_left, points[i][0])
return res
复杂度分析
时间复杂度:O($nlogn$) 空间复杂度:O(1)
Q2. LeetCode 435
与Q1类似,但注意最后结果需要减去找到的非重合区间数量
class Solution:
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
if len(intervals) == 0:
return 0
intervals.sort(key=lambda x:x[1])
count = 1
curr_right = intervals[0][1]
for i in range(1, len(intervals)):
if intervals[i][0] >= curr_right:
count += 1
curr_right = intervals[i][1]
else:
curr_right = min(intervals[i][1], curr_right)
return len(intervals) - count
复杂度分析
时间复杂度:O($nlogn$) 空间复杂度:O($n$)
Q3. LeetCode 763
可以用数组替代字典,降低空间开销
class Solution:
def partitionLabels(self, s: str) -> List[int]:
hashmap = {}
for i in range(len(s)):
hashmap[s[i]] = i
res = []
end = 0
start = 0
for i in range(len(s)):
# 先求最大值,再考虑是否匹配(特殊情况:第一个substring只含有首字母)
end = max(end, hashmap[s[i]])
if i == end:
res.append(end - start + 1)
start = i + 1
return res
复杂度分析
时间复杂度:O($n$) 空间复杂度:O($n$)
