435. 无重叠区间 

class Solution:def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:intervals.sort(key=lambda x:x[0])result = 0for i in range(1,len(intervals)):if intervals[i][0] < intervals[i-1][1]:result += 1intervals[i][1] = min(intervals[i-1][1],intervals[i][1]) #这里注意要取最小值return result

class Solution:def partitionLabels(self, s: str) -> List[int]:#统计每个字母出现的最后位置last_occurrence = {}for i,ch in enumerate(s):last_occurrence[ch] = iresult = []start = 0 end = 0for i,ch in enumerate(s):end = max(end,last_occurrence[ch])if i == end :result.append(end-start+1)start = i+1return result

 enumerate用法：

对一个列表，既要遍历索引又要遍历元素时可以使用enumerate函数：

list1 = ["这", "是", "一个", "测试"]
for index, item in enumerate(list1):print(index,item)

 enumerate还可以接收第二个参数，用于指定索引起始值

list1 = ["这", "是", "一个", "测试"]
for index, item in enumerate(list1,1):print(index,item)

class Solution:def merge(self, intervals: List[List[int]]) -> List[List[int]]:intervals.sort(key = lambda x:x[0])result = []for i in range(1,len(intervals)):if intervals[i][0] > intervals[i-1][1]:result.append(intervals[i-1])else:intervals[i][0] = intervals[i-1][0]intervals[i][1] = max(intervals[i-1][1],intervals[i][1])if i == len(intervals)-1:result.append(intervals[i])return result

或者先把第一个元素添加进result，接着进行比较

class Solution:def merge(self, intervals):result = []if len(intervals) == 0:return result  # 区间集合为空直接返回intervals.sort(key=lambda x: x[0])  # 按照区间的左边界进行排序result.append(intervals[0])  # 第一个区间可以直接放入结果集中for i in range(1, len(intervals)):if result[-1][1] >= intervals[i][0]:  # 发现重叠区间# 合并区间，只需要更新结果集最后一个区间的右边界，因为根据排序，左边界已经是最小的result[-1][1] = max(result[-1][1], intervals[i][1])else:result.append(intervals[i])  # 区间不重叠return result