435. 无重叠区间
思路：和最少数量引爆气球的箭的思路基本都是一致了！贪心就是比较左边的值是否大于下一个右边的值
class Solution:def eraseOverlapIntervals(self, points: List[List[int]]) -> int:points.sort(key=lambda x: (x[0], x[1]))# 比较边界res = points[0][1]count = 0for i in range(1, len(points)):if points[i][0] < res:res = min(res, points[i][1])count += 1else:res = points[i][1]return count763. 划分字母区间
贪心思路：第一个字母最后出现的位置的过程中其他字母也出现的最后位置的最大值，就是一次划分区域了！
class Solution:def partitionLabels(self, s: str) -> List[int]:# 字母出现次数？# 每一次字母最后出现的位置以及mm = {}for i in range(len(s)):mm[s[i]] = istart = 0max_temp = mm[s[0]]res = []for i in range(len(s)):if max_temp == i:res.append(max_temp - start + 1)start = i + 1if i + 1 < len(s):max_temp = mm[s[i + 1]]elif mm[s[i]] > max_temp:max_temp = mm[s[i]]return resclass Solution:def partitionLabels(self, s: str) -> List[int]:last_occurrence = {} # 存储每个字符最后出现的位置for i, ch in enumerate(s):last_occurrence[ch] = iresult = []start = 0end = 0for i, ch in enumerate(s):end = max(end, last_occurrence[ch]) # 找到当前字符出现的最远位置if i == end: # 如果当前位置是最远位置，表示可以分割出一个区间result.append(end - start + 1)start = i + 1return result56. 合并区间
思路：就是如果区间的重叠的哇，就去重叠区间之间的最小值和最大值!关键还是拆分排序
class Solution:def merge(self, intervals: List[List[int]]) -> List[List[int]]:res = []intervals.sort(key=lambda x :(x[0], x[1]))index = 0max_broad = intervals[0][1]for i in range(1, len(intervals)):if intervals[i][0] <= max_broad:intervals[index][0] = min(intervals[index][0], intervals[i][0])intervals[index][1] = max(intervals[index][1], intervals[i][1])max_broad = intervals[index][1]else:res.append(intervals[index])index = imax_broad = intervals[i][1]if intervals[index] not in res:res.append(intervals[index])return res
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