代码随想录算法训练营day36|435. 无重叠区间、763.划分字母区间、56. 合并区间

435. 无重叠区间 

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort(key=lambda x:x[0])
        result = 0

        for i in range(1,len(intervals)):
            if intervals[i][0] < intervals[i-1][1]:
                result += 1
                intervals[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] = i
        
        result = []
        start = 0 
        end = 0

        for i,ch in enumerate(s):
            end = max(end,last_occurrence[ch])
            if i == end :
                result.append(end-start+1)
                start = i+1
        
        return 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