LEETCODE-DAY36

---

title: LEETCODE-DAY36
date: 2024-03-27 18:49:59
tags:

今日内容：435. 无重叠区间、763.划分字母区间、56. 合并区间

T1

思路类似DAY35 T3

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort()
        j=0
        for i in range(len(intervals)):
            if i>0 and intervals[i][0]<intervals[i-1][1]:
                #因为要使j尽可能小，故把较大的上界对应的区间remove
                intervals[i][1]=min(intervals[i][1],intervals[i-1][1])
                j+=1
        return j

AC

T2

class 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] = i

        result = []
        start = 0
        end = 0
        for 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 + 1

        return result

T3

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
            res=list()
            intervals.sort()
            for i in range(len(intervals)):
                if i>0 and intervals[i][0]<=intervals[i-1][1]:
                    intervals[i][0]=intervals[i-1][0]
                    intervals[i][1]=max(intervals[i][1],intervals[i-1][1])
                elif i>0 and intervals[i][0]>intervals[i-1][1]:
                    res.append(intervals[i-1])

            return res

输入
intervals =
[[1,3],[2,6],[8,10],[15,18]]
输出
[[1,6],[8,10]]
预期结果
[[1,6],[8,10],[15,18]]

输入
intervals =
[[1,4],[4,5]]
输出
[]
预期结果
[[1,5]]

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
            res=list()
            intervals.sort()
            for i in range(len(intervals)):
                if i>0 and intervals[i][0]<=intervals[i-1][1]:
                    intervals[i][0]=intervals[i-1][0]
                    intervals[i][1]=max(intervals[i][1],intervals[i-1][1])
                elif i>0 and intervals[i][0]>intervals[i-1][1]:
                    res.append(intervals[i-1])
            res.append(intervals[i])

            return res

AC