一个方法解决三道区间问题

1288. 删除被覆盖区间
56. 合并区间
986. 区间列表的交集

# 1288. 删除被覆盖区间
class Solution:def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:# 按照起点升序排列，起点相同时，按照终点降序排列intervals.sort(key = lambda x: (x[0], -x[1]))# 记录被覆盖区间的个数res = 0# 记录合并区间的起点和终点left, right = intervals[0][0], intervals[0][1]i = 1while i < len(intervals):curr = intervals[i]# 覆盖if left <= curr[0] and curr[1] <= right:res += 1# 合并if curr[0] <= right <= curr[1]:right = curr[1]# 无交集，更新left rightif right <= curr[0]:left, right = curr[0], curr[1]i += 1return len(intervals) - res

# 56. 合并区间
class Solution:def merge(self, intervals: List[List[int]]) -> List[List[int]]:intervals.sort()res = [intervals[0]]i = 1while i < len(intervals):last = res[-1]curr = intervals[i]if curr[0] <= last[1]:last[1] = max(last[1], curr[1])else:res.append(curr)i += 1return res

# 986. 区间列表的交集
class Solution:def intervalIntersection(self, firstList: List[List[int]], secondList: List[List[int]]) -> List[List[int]]:i, j = 0, 0res = []while i < len(firstList) and j < len(secondList):a1, a2 = firstList[i][0], firstList[i][1]b1, b2 = secondList[j][0], secondList[j][1]if b2>=a1 and b1<=a2:c1, c2 = max(a1, b1), min(a2, b2)res.append([c1, c2])if a2 < b2:i += 1else:j += 1return res