Leetcode Insert Intervals_intervals insert-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_39544046/article/details/105942487

本文介绍了一种在已排序的非重叠区间中插入新区间并进行必要合并的算法实现。通过两种方法，一种返回新的合并后的区间列表，另一种在原地修改输入区间列表。这两种方法都详细展示了如何处理新区间与已有区间之间的各种相交情况。

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Ex.2: Insert Intervals

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.
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version 1:

def insert(intervals, newInterval):
    merged = []
    for i in intervals:
        if newInterval is None or i[1] < newInterval[0]:
            merged += i,
        elif i[0] > newInterval[1]:
            merged += newInterval,
            merged += i,
            newInterval = None
        else:
            newInterval[0] = min(newInterval[0], i[0])
            newInterval[1] = max(newInterval[1], i[1])
    if newInterval is not None:
        merged += newInterval,
    return merged

intervals = [[1,2],[3,4],[6,7],[7,8]]
newInterval = [8,9]
insert(intervals, newInterval)

in place:

def insert(intervals, newInterval):
    for i in range(len(intervals)):
        # 根据左区间找到对应插入位置
        if intervals[i][0] > newInterval[0]:
            # 判断新区间与左侧区间的关系
            # 相离
            if intervals[i-1][1] < newInterval[0]:
                intervals.insert(i, newInterval)
                i = i + 1
            # 相交
            else:
                # 相交不包含
                if intervals[i-1][1] < newInterval[1]:
                    intervals[i-1][1] = newInterval[1]
                # 包含不需要任何操作
            
            # 判断新区间与插入位置右侧区间的情况
            # 相交则完成合并
            if intervals[i][0] <= newInterval[1]:
                intervals[i-1][1] = intervals[i][1]
                intervals.remove(intervals[i])
            # 插入操作完成后，推出循环
            break
            
            # 如果插入的位置在最右侧
        if i == len(intervals) - 1:
            # 判断最后一个区间与新区间的关系
            # 相离
            if intervals[i][1] < newInterval[0]:
                intervals.append(newInterval)
            else:
                intervals[i][1] = newInterval[1]
                    
    return intervals
        
                
intervals = [[1,2],[3,4],[6,7],[7,8]]
newInterval = [8,9]
insert(intervals, newInterval)